What Is Physics?
Human constructions are supposed to be stable in spite of the forces that act on them.
A building, for example, should be stable in spite of the gravitational force and wind
forces on it, and a bridge should be stable in spite of the gravitational force pulling it
downward and the repeated jolting it receives from cars and trucks.
One focus of physics is on what allows an object to be stable in spite of any
forces acting on it. In this chapter we examine the two main aspects of stability:
the equilibrium of the forces and torques acting on rigid objects and the elasticity
of nonrigid objects, a property that governs how such objects can deform. When
this physics is done correctly, it is the subject of countless articles in physics and
engineering journals; when it is done incorrectly, it is the subject of countless
articles in newspapers and legal journals.
Equilibrium
Consider these objects: (1) a book resting on a table, (2) a hockey puck sliding
with constant velocity across a frictionless surface, (3) the rotating blades of a
ceiling fan, and (4) the wheel of a bicycle that is traveling along a straight path at
constant speed. For each of these four objects,
CHAPTER 12
Equilibrium and Elasticity
12-1 EQUILIBRIUM
After reading this module, you should be able to . . .
12.01 Distinguish between equilibrium and static equilibrium.
12.02 Specify the four conditions for static equilibrium.
12.03 Explain center of gravity and how it relates to center of
mass.
12.04 For a given distribution of particles, calculate the coor-
dinates of the center of gravity and the center of mass.
Key Ideas
Learning Objectives
327327
●A rigid body at rest is said to be in static equilibrium. For
such a body, the vector sum of the external forces acting on it
is zero:
(balance of forces).
If all the forces lie in the xy plane, this vector equation is
equivalent to two component equations:
Fnet,x0and Fnet,y0(balance of forces).
●Static equilibrium also implies that the vector sum of the
external torques acting on the body about any point is zero, or
(balance of torques).t
:
net 0
F
:
net 0
If the forces lie in the xy plane, all torque vectors are parallel
to the zaxis, and the balance-of-torques equation is equiva-
lent to the single component equation
tnet,z0(balance of torques).
●The gravitational force acts individually on each element of
a body. The net effect of all individual actions may be found by
imagining an equivalent total gravitational force acting at
the center of gravity. If the gravitational acceleration is the
same for all the elements of the body, the center of gravity is
at the center of mass.
g
:
F
:
g
1. The linear momentum of its center of mass is constant.
2. Its angular momentum about its center of mass, or about any other point, is
also constant.
We say that such objects are in equilibrium. The two requirements for
equilibrium are then
.(12-1)
Our concern in this chapter is with situations in which the constants in
Eq. 12-1 are zero; that is, we are concerned largely with objects that are not mov-
ing in any way—either in translation or in rotation—in the reference frame
from which we observe them. Such objects are in static equilibrium. Of the four
objects mentioned near the beginning of this module, only one—the book rest-
ing on the table—is in static equilibrium.
The balancing rock of Fig. 12-1 is another example of an object that, for the
present at least, is in static equilibrium. It shares this property with countless
other structures, such as cathedrals, houses, filing cabinets, and taco stands, that
remain stationary over time.
As we discussed in Module 8-3, if a body returns to a state of static equilib-
rium after having been displaced from that state by a force, the body is said to be
in stable static equilibrium. A marble placed at the bottom of a hemispherical
bowl is an example. However, if a small force can displace the body and end the
equilibrium, the body is in unstable static equilibrium.
A Domino. For example, suppose we balance a domino with the domino’s
center of mass vertically above the supporting edge, as in Fig. 12-2a.The torque
about the supporting edge due to the gravitational force F
:
gon the domino is
zero because the line of action of F
:
gis through that edge.Thus, the domino is in
equilibrium. Of course, even a slight force on it due to some chance distur-
bance ends the equilibrium. As the line of action of F
:
gmoves to one side of
the supporting edge (as in Fig. 12-2b), the torque due to F
:
gincreases the rota-
tion of the domino. Therefore, the domino in Fig. 12-2ais in unstable static
equilibrium.
The domino in Fig. 12-2cis not quite as unstable. To topple this domino,
a force would have to rotate it through and then beyond the balance position of
Fig. 12-2a, in which the center of mass is above a supporting edge. A slight force
will not topple this domino, but a vigorous flick of the finger against the domino
certainly will. (If we arrange a chain of such upright dominos, a finger flick
against the first can cause the whole chain to fall.)
A Block. The child’s square block in Fig. 12-2dis even more stable because
its center of mass would have to be moved even farther to get it to pass above a
supporting edge. A flick of the finger may not topple the block. (This is why you
P
:a constant and L
:a constant
L
:
P
:
328 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
Figure 12-1 A balancing rock.Although its
perch seems precarious, the rock is in static
equilibrium.
Kanwarjit Singh Boparai/Shutterstock
Figure 12-2 (a) A domino balanced on one
edge,with its center of mass vertically above
that edge.The gravitational force F
:
gon the
domino is directed through the supporting
edge.(b) If the domino is rotated even slightly
from the balanced orientation,then causes
a torque that increases the rotation. (c) A
domino upright on a narrow side is
somewhat more stable than the domino in
(a). (d) A square block is even more stable.
F
:
g
B
B
com
Supporting
edge
(a) (b) (c) (d)
FgFgFg
Fg
To tip the block, the center of mass must
pass over the supporting edge.
never see a chain of toppling square blocks.) The worker in Fig. 12-3 is like both
the domino and the square block: Parallel to the beam, his stance is wide and he is
stable; perpendicular to the beam, his stance is narrow and he is unstable (and at
the mercy of a chance gust of wind).
The analysis of static equilibrium is very important in engineering practice.The
design engineer must isolate and identify all the external forces and torques that
may act on a structure and, by good design and wise choice of materials, ensure that
the structure will remain stable under these loads. Such analysis is necessary to en-
sure, for example, that bridges do not collapse under their traffic and wind loads and
that the landing gear of aircraft will function after the shock of rough landings.
The Requirements of Equilibrium
The translational motion of a body is governed by Newton’s second law in its
linear momentum form, given by Eq. 9-27 as
(12-2)
If the body is in translational equilibrium—that is, if is a constant—then
0 and we must have
(balance of forces). (12-3)
The rotational motion of a body is governed by Newton’s second law in its
angular momentum form, given by Eq. 11-29 as
(12-4)
If the body is in rotational equilibrium—that is, if is a constant—then 0
and we must have
(balance of torques). (12-5)
Thus, the two requirements for a body to be in equilibrium are as follows:
t
:
net 0
dL
:/dtL
:
t
:
net dL
:
dt .
F
:
net 0
dP
:/dt
P
:
F
:
net dP
:
dt .
329
12-1 EQUILIBRIUM
Figure 12-3 A construction worker balanced
on a steel beam is in static equilibrium but
is more stable parallel to the beam than per-
pendicular to it.
Robert Brenner/PhotoEdit
1. The vector sum of all the external forces that act on the body must be zero.
2. The vector sum of all external torques that act on the body, measured about any
possible point, must also be zero.
These requirements obviously hold for static equilibrium. They also hold for the
more general equilibrium in which and are constant but not zero.
Equations 12-3 and 12-5, as vector equations, are each equivalent to three
independent component equations, one for each direction of the coordinate axes:
Balance of Balance of
forces torques
Fnet,x0tnet,x0
Fnet,y0tnet,y0 (12-6)
Fnet,z0tnet,z0
The Main Equations. We shall simplify matters by considering only situations in
which the forces that act on the body lie in the xy plane. This means that the only
torques that can act on the body must tend to cause rotation around an axis parallel to
L
:
P
:
the zaxis.With this assumption, we eliminate one force equation and two torque equa-
tions from Eqs.12-6,leaving
Fnet,x0(balance of forces), (12-7)
Fnet,y0(balance of forces), (12-8)
tnet,z0(balance of torques). (12-9)
Here, tnet,zis the net torque that the external forces produce either about the
zaxis or about any axis parallel to it.
A hockey puck sliding at constant velocity over ice satisfies Eqs. 12-7, 12-8,
and 12-9 and is thus in equilibrium but not in static equilibrium. For static equilib-
rium, the linear momentum of the puck must be not only constant but also
zero; the puck must be at rest on the ice. Thus, there is another requirement for
static equilibrium:
P
:
330 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
3. The linear momentum of the body must be zero.P
:
Checkpoint 1
The figure gives six overhead views of a uniform rod on which two or more forces act
perpendicularly to the rod.If the magnitudes of the forces are adjusted properly (but kept
nonzero),in which situations can the rod be in static equilibrium?
(a) (b) (c)
(d) (e) (f)
The Center of Gravity
The gravitational force on an extended body is the vector sum of the gravita-
tional forces acting on the individual elements (the atoms) of the body. Instead of
considering all those individual elements, we can say that
The gravitational force on a body effectively acts at a single point, called the
center of gravity (cog) of the body.
F
:
g
Here the word “effectively” means that if the gravitational forces on the individual
elements were somehow turned off and the gravitational force at the center of
gravity were turned on, the net force and the net torque (about any point) acting on
the body would not change.
Until now, we have assumed that the gravitational force acts at the center
of mass (com) of the body.This is equivalent to assuming that the center of grav-
ity is at the center of mass. Recall that, for a body of mass M, the force is equal
to , where is the acceleration that the force would produce if the body wereg
:
Mg
:
F
:
g
F
:
g
F
:
g
Proof
First, we consider the individual elements of the body. Figure 12-4ashows an
extended body, of mass M, and one of its elements, of mass mi. A gravitational
force acts on each such element and is equal to The subscript on
means is the gravitational acceleration at the location of the element i (it can be
different for other elements).
For the body in Fig. 12-4a, each force F
:
gi acting on an element produces a
torque tion the element about the origin O, with a moment arm xi. Using Eq. 10-
41 (trF) as a guide, we can write each torque tias
tixiFgi. (12-10)
The net torque on all the elements of the body is then
tnet tixiFgi. (12-11)
Next, we consider the body as a whole. Figure 12-4bshows the gravitational
force acting at the body’s center of gravity. This force produces a torque ton
the body about O, with moment arm xcog.Again using Eq. 10-41, we can write this
torque as
txcogFg. (12-12)
The gravitational force F
:
gon the body is equal to the sum of the gravitational
forces F
:
gi on all its elements, so we can substitute Fgi for Fgin Eq. 12-12 to write
txcog Fgi. (12-13)
Now recall that the torque due to force F
:
gacting at the center of gravity
is equal to the net torque due to all the forces F
:
gi acting on all the elements of
the body. (That is how we defined the center of gravity.) Thus, tin Eq. 12-13 is
equal to tnet in Eq. 12-11. Putting those two equations together, we can write
xcog Fgi xiFgi.
Substituting migifor Fgi gives us
xcog migiximigi. (12-14)
Now here is a key idea: If the accelerations giat all the locations of the elements
are the same, we can cancel gifrom this equation to write
xcog miximi. (12-15)
The sum miof the masses of all the elements is the mass Mof the body.
Therefore,we can rewrite Eq.12-15 as
(12-16)xcog 1
M
ximi.
F
:
g
g
:
i
g
:
i
mig
:
i.F
:
gi
331
12-1 EQUILIBRIUM
If is the same for all elements of a body, then the body’s center of gravity (cog)
is coincident with the body’s center of mass (com).
g
:
This is approximately true for everyday objects because varies only a little
along Earth’s surface and decreases in magnitude only slightly with altitude.
Thus, for objects like a mouse or a moose, we have been justified in assuming that
the gravitational force acts at the center of mass. After the following proof, we
shall resume that assumption.
g
:
Ox
y
xcog
cog
(b)
Line of
action
Moment
arm
Fg
Ox
y
mi
xi
(a)
Line of
action
Fgi
Moment
arm
Figure 12-4 (a) An element of mass miin an
extended body.The gravitational force F
:
gi
on the element has moment arm xiabout
the origin Oof the coordinate system. (b)
The gravitational force F
:
gon a body is said
to act at the center of gravity (cog) of the
body. Here F
:
ghas moment arm xcog about
origin O.
to fall freely. In the proof that follows, we show that
Some Examples of Static Equilibrium
Here we examine several sample problems involving static equilibrium. In
each, we select a system of one or more objects to which we apply the equa-
tions of equilibrium (Eqs. 12-7, 12-8, and 12-9).The forces involved in the equi-
librium are all in the xy plane, which means that the torques involved are par-
allel to the zaxis. Thus, in applying Eq. 12-9, the balance of torques, we select
an axis parallel to the zaxis about which to calculate the torques. Although
Eq. 12-9 is satisfied for any such choice of axis, you will see that certain choices
simplify the application of Eq. 12-9 by eliminating one or more unknown force
terms.
332 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
12-2 SOME EXAMPLES OF STATIC EQUILIBRIUM
Learning Objectives
origin (about which to calculate torques) can simplify the
calculations by eliminating one or more unknown forces
from the torque equation.
●A rigid body at rest is said to be in static equilibrium. For
such a body, the vector sum of the external forces acting on it
is zero:
(balance of forces).
If all the forces lie in the xy plane, this vector equation is
equivalent to two component equations:
Fnet,x0and Fnet,y0(balance of forces).
F
:
net 0
●Static equilibrium also implies that the vector sum of the
external torques acting on the body about any point is zero, or
(balance of torques).
If the forces lie in the xy plane, all torque vectors are parallel
to the zaxis, and the balance-of-torques equation is equiva-
lent to the single component equation
tnet,z0(balance of torques).
t
:net 0
After reading this module, you should be able to . . .
12.05 Apply the force and torque conditions for static
equilibrium.
12.06 Identify that a wise choice about the placement of the
Key Ideas
Checkpoint 2
The figure gives an overhead view of a uniform rod in static equilibrium. (a) Can you
find the magnitudes of unknown forces and by balancing the forces? (b) If you
wish to find the magnitude of force by using a balance of torques equation, where
should you place a rotation axis to eliminate from the equation? (c) The magnitude
of turns out to be 65 N.What then is the magnitude of ?
F
:
1
F
:
2
F
:
1
F
:
2
F
:
2
F
:
1
20 N dd2d4d
10 N 30 N
F2
F1
The right side of this equation gives the coordinate xcom of the body’s center of
mass (Eq. 9-4).We now have what we sought to prove. If the acceleration of grav-
ity is the same at all locations of the elements in a body, then the coordinates of
the body’s com and cog are identical:
xcog xcom. (12-17)
333
12-2 SOME EXAMPLES OF STATIC EQUILIBRIUM
Additional examples, video, and practice available at WileyPLUS
which gives us
15.44 N 15 N. (Answer)
Now, solving Eq.12-18 for Fland substituting this result, we find
Fl(Mm)gFr
(2.7 kg 1.8 kg)(9.8 m/s2)15.44 N
28.66 N 29 N. (Answer)
Notice the strategy in the solution: When we wrote an equa-
tion for the balance of force components, we got stuck with two
unknowns. If we had written an equation for the balance of
torques around some arbitrary axis, we would have again got-
ten stuck with those two unknowns. However, because we
chose the axis to pass through the point of application of one of
the unknown forces, here l, we did not get stuck. Our choice
neatly eliminated that force from the torque equation,allowing
us to solve for the other unknown force magnitude Fr.Then we
returned to the equation for the balance of force components
to find the remaining unknown force magnitude.
F
:
1
4(2.7 kg)(9.8 m/s2)1
2(1.8 kg)(9.8 m/s2)
Fr1
4Mg 1
2mg
In Fig. 12-5a, a uniform beam, of length Land mass
m1.8 kg, is at rest on two scales. A uniform block, with
mass M2.7 kg, is at rest on the beam, with its center a dis-
tance L/4 from the beam’s left end.What do the scales read?
KEY IDEAS
The first steps in the solution of any problem about static equi-
librium are these: Clearly define the system to be analyzed
and then draw a free-body diagram of it, indicating all the
forces on the system. Here, let us choose the system as the
beam and block taken together.Then the forces on the system
are shown in the free-body diagram of Fig. 12-5b. (Choosing
the system takes experience, and often there can be more than
one good choice.) Because the system is in static equilibrium,
we can apply the balance of forces equations (Eqs. 12-7 and
12-8) and the balance of torques equation (Eq.12-9) to it.
Calculations: The normal forces on the beam from the scales
are F
:
lon the left and F
:
ron the right.The scale readings that we
want are equal to the magnitudes of those forces. The gravita-
tional force F
:
g,beam on the beam acts at the beam’s center of
mass and is equal to Similarly,the gravitational force F
:
g,block
on the block acts at the block’s center of mass and is equal to
However, to simplify Fig. 12-5b, the block is repre-
sented by a dot within the boundary of the beam and vector
F
:
g,block is drawn with its tail on that dot. (This shift of the
vector F
:
g,block along its line of action does not alter the
torque due to F
:
g,block about any axis perpendicular to the figure.)
The forces have no xcomponents, so Eq. 12-7
(Fnet,x0) provides no information. For the ycomponents,
Eq. 12-8 (Fnet,y0) gives us
FlFrMg mg 0. (12-18)
This equation contains two unknowns, the forces Fl
and Fr, so we also need to use Eq. 12-9, the balance of
torques equation. We can apply it to any rotation axis per-
pendicular to the plane of Fig. 12-5. Let us choose a rota-
tion axis through the left end of the beam. We shall also
use our general rule for assigning signs to torques: If a
torque would cause an initially stationary body to rotate
clockwise about the rotation axis, the torque is negative. If
the rotation would be counterclockwise, the torque is posi-
tive. Finally, we shall write the torques in the form rF,
where the moment arm ris 0 for L/4 for L/2 for
and Lfor .
We now can write the balancing equation (tnet,z0) as
(0)(Fl)(L/4)(Mg)(L/2)(mg)(L)(Fr)0,
F
:
r
mg
:
,Mg
:
,Fl
:,
Mg
:
.
mg
:
.Figure 12-5 (a) A beam of mass msupports a block of mass M.
(b) A free-body diagram, showing the forces that act on the
system beam block.
y
x
(b)
Block Beam
L
4
L
2
(a)
Block Beam
Scale Scale
m
M
L
L
4
S
ystem
Fl
Fr
Fg,beam = mg
Fg,block = Mg
The vertical forces balance
but that is not enough.
We must also balance
torques, with a wise
choice of rotation axis.
Sample Problem 12.01 Balancing a horizontal beam
334 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
magnitude Fof the net force. Because we know Tc, we apply
the force balancing equations to the beam.
Calculations: For the horizontal balance, we can rewrite
Fnet,x0 as
FhTc0, (12-20)
and so FhTc6093 N.
For the vertical balance, we write Fnet,y0 as
Fvmg Tr0.
Substituting Mg for Trand solving for Fv, we find that
Fv(mM)g(85 kg 430 kg)(9.8 m/s2)
5047 N.
From the Pythagorean theorem, we now have
(Answer)
Note that Fis substantially greater than either the combined
weights of the safe and the beam, 5000 N, or the tension in
the horizontal wire, 6100 N.
2(6093 N)2(5047 N)27900 N.
F2F2
hFv
2
Sample Problem 12.02 Balancing a leaning boom
Figure 12-6ashows a safe (mass M430 kg) hanging by a
rope (negligible mass) from a boom (a1.9 m and b
2.5 m) that consists of a uniform hinged beam (m85 kg)
and horizontal cable (negligible mass).
(a) What is the tension Tcin the cable? In other words, what is
the magnitude of the force on the beam from the cable?
KEY IDEAS
The system here is the beam alone, and the forces on it are
shown in the free-body diagram of Fig. 12-6b. The force
from the cable is . The gravitational force on the beam
acts at the beam’s center of mass (at the beam’s center) and
is represented by its equivalent The vertical component
of the force on the beam from the hinge is , and the hori-
zontal component of the force from the hinge is . The
force from the rope supporting the safe is . Because beam,
rope, and safe are stationary, the magnitude of is equal toT
:
r
T
:
r
F
:
h
F
:
v
mg
:
.
T
:
c
T
:
c
Additional examples, video, and practice available at WileyPLUS
the weight of the safe: TrMg. We place the origin Oof an
xy coordinate system at the hinge. Because the system is in
static equilibrium, the balancing equations apply to it.
Calculations: Let us start with Eq. 12-9 (tnet,z0). Note
that we are asked for the magnitude of force and not of
forces and acting at the hinge, at point O. To eliminate
and from the torque calculation, we should calculate
torques about an axis that is perpendicular to the figure at
point O.Then and will have moment arms of zero. The
lines of action for , , and are dashed in Fig. 12-6b.
The corresponding moment arms are a,b, and b/2.
Writing torques in the form of rFand using our rule about
signs for torques,the balancing equation tnet,z0 becomes
(12-19)
Substituting Mg for Trand solving for Tc, we find that
6093 N 6100 N. (Answer)
(b) Find the magnitude Fof the net force on the beam from
the hinge.
KEY IDEA
Now we want the horizontal component Fhand vertical
component Fvso that we can combine them to get the
(9.8 m/s2)(2.5 m)(430 kg 85/2 kg)
1.9 m
Tcgb(M1
2m)
a
(a)(Tc)(b)(Tr)(1
2b)(mg)0.
mg
:
T
:
r
T
:
c
F
:
v
F
:
h
F
:
v
F
:
h
F
:
v
F
:
h
T
:
c
Figure 12-6 (a)
A heavy safe is
hung from a
boom consisting
of a horizontal
steel cable and a
uniform beam.
(b) A free-body
diagram for the
beam.
b
a
Cable
Rope
Hinge
Beam com
θ
m
(a)
y
x
Beam
mg
(b)
θ
O
M
Tc
Tr
Fv
Fh
Here is the
wise choice of
rotation axis.
335
12-2 SOME EXAMPLES OF STATIC EQUILIBRIUM
line extending through the force vector), so that its tail is on
the dot. (The shift does not alter a torque due to about
any axis perpendicular to the figure.Thus, the shift does not af-
fect the torque balancing equation that we shall be using.)
The only force on the ladder from the wall is the hori-
zontal force (there cannot be a frictional force along a
frictionless wall, so there is no vertical force on the ladder
from the wall). The force F
:
pon the ladder from the pave-
ment has two components: a horizontal component F
:
px that
is a static frictional force and a vertical component F
:
py that is
a normal force.
To apply the balancing equations, let’s start with the
torque balancing of Eq. 12-9 (tnet,z0). To choose an axis
about which to calculate the torques, note that we have
unknown forces ( and F
:
p) at the two ends of the ladder.To
eliminate, say, F
:
pfrom the calculation, we place the axis at
point O, perpendicular to the figure (Fig. 12-7b). We also
place the origin of an xy coordinate system at O.We can find
torques about Owith any of Eqs. 10-39 through 10-41, but
Eq. 10-41 (trF) is easiest to use here. Making a wise
choice about the placement of the origin can make our torque
calculation much easier.
To find the moment arm rof the horizontal force w
from the wall, we draw a line of action through that vector
F
:
F
:
w
F
:
w
Mg
:
Sample Problem 12.03 Balancing a leaning ladder
In Fig. 12-7a, a ladder of length L12 m and mass m
45 kg leans against a slick wall (that is, there is no friction
between the ladder and the wall). The ladder’s upper end is
at height h9.3 m above the pavement on which the lower
end is supported (the pavement is not frictionless). The
ladder’s center of mass is L/3 from the lower end, along the
length of the ladder.A firefighter of mass M72 kg climbs
the ladder until her center of mass is L/2 from the lower
end. What then are the magnitudes of the forces on the lad-
der from the wall and the pavement?
KEY IDEAS
First, we choose our system as being the firefighter and lad-
der, together, and then we draw the free-body diagram of
Fig. 12-7bto show the forces acting on the system. Because
the system is in static equilibrium, the balancing equations
for both forces and torques (Eqs. 12-7 through 12-9) can be
applied to it.
Calculations: In Fig.12-7b, the firefighter is represented with
a dot within the boundary of the ladder. The gravitational
force on her is represented with its equivalent expression
and that vector has been shifted along its line of action (the
Mg
:
,
A
System
Fire-
fighter
com
Ladder
com
Frictionless
h
L
(a)
Fire-
fighter
Ladder
a/2
(b)
y
x
mg
Mg
O
a/3
a
Fpx
Fpy
FwHere are all
the forces.
Figure 12-7 (a) A firefighter climbs halfway up a ladder that is leaning against a frictionless
wall.The pavement beneath the ladder is not frictionless. (b) A free-body diagram, showing
the forces that act on the firefighter ladder system.The origin Oof a coordinate system is
placed at the point of application of the unknown force F
:
p(whose vector components F
:
px and
F
:
py are shown). (Figure 12-7 continues on following page.)
Additional examples, video, and practice available at WileyPLUS
336 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
Figure 12-7 (Continued from previous page)
(c) Calculating the torques. (d) Balancing
the forces. In WileyPLUS,this figure is
available as an animation with voiceover.
(it is the horizontal dashed line shown in Fig. 12-7c). Then r
is the perpendicular distance between Oand the line of ac-
tion. In Fig. 12-7c,rextends along the yaxis and is equal to
the height h.We similarly draw lines of action for the gravita-
tional force vectors and and see that their moment
arms extend along the xaxis. For the distance ashown in Fig.
12-7a, the moment arms are a/2 (the firefighter is halfway up
the ladder) and a/3 (the ladder’s center of mass is one-third of
the way up the ladder), respectively.The moment arms for
and are zero because the forces act at the origin.Fpy
:Fpx
:
mg
:
Mg
:
Then Eq. 12-21 gives us
407 N 410 N. (Answer)
Now we need to use the force balancing equations and
Fig. 12-7d.The equation Fnet,x0 gives us
FwFpx 0,
so Fpx Fw410 N. (Answer)
The equation Fnet,y0 gives us
Fpy Mg mg 0,
so Fpy (Mm)g(72 kg 45 kg)(9.8 m/s2)
1146.6 N 1100 N. (Answer)
(9.8 m/s2)(7.58 m)(72/2 kg 45/3 kg)
9.3 m
Fwga(M/2 m/3)
h
Now, with torques written in the form rF, the balancing
equation tnet,z0 becomes
(h)(Fw)(a/2)(Mg)(a/3)(mg)
(0)(Fpx)(0)(Fpy)0. (12-21)
(A positive torque corresponds to counterclockwise rotation
and a negative torque corresponds to clockwise rotation.)
Using the Pythagorean theorem for the right triangle
made by the ladder in Fig.11-7a, we find that
.a1L2h27.58 m
y
x
mg
Mg
O
Fpy
y
x
O
Fpx
Fw
(c)
(d)
y
x
O
Fpx
Fpy
y
x
O
h
Fw
y
x
Mg
O
a/2a/3
y
x
mg
O
Choosing the
rotation axis
here eliminates
the torques
due to these
forces.
This moment
arm is
perpendicular
to the
line of action.
These horizontal
forces balance.
These
vertical
forces
balance.
Here
too.
Here
too.
337
12-2 SOME EXAMPLES OF STATIC EQUILIBRIUM
which gives us
Dividing this new result for the normal force at the right by
the original result and then substituting for d, we obtain
Substituting the values of h60 m, R9.8 m, and u5.5
leads to
Thus, our simple model predicts that, although the tilt is
modest, the normal force on the tower’s southern wall has
increased by about 30%. One danger to the tower is that the
force may cause the southern wall to buckle and explode
outward. The cause of the leaning is the compressible soil
beneath the tower, which worsened with each rainfall.
Recently engineers have stabilized the tower and partially
reversed the leaning by installing a drainage system.
F
NR
FNR
1.29.
F
NR
FNR
Rd
R1d
R10.5h tan u
R.
F
NR (Rd)
2Rmg.
Sample Problem 12.04 Balancing the leaning Tower of Pisa
Let’s assume that the Tower of Pisa is a uniform hollow
cylinder of radius R9.8 m and height h60 m. The
center of mass is located at height h/2, along the cylin-
der’s central axis. In Fig. 12-8a, the cylinder is upright. In
Fig. 12-8b, it leans rightward (toward the tower’s south-
ern wall) by u5.5, which shifts the com by a distance d.
Let’s assume that the ground exerts only two forces on
the tower. A normal force acts on the left (northern)
wall, and a normal force acts on the right (southern)
wall. By what percent does the magnitude FNR increase
because of the leaning?
KEY IDEA
Because the tower is still standing, it is in equilibrium and
thus the sum of torques calculated around any point must
be zero.
Calculations: Because we want to calculate FNR on the
right side and do not know or want FNL on the left side, we
use a pivot point on the left side to calculate torques. The
forces on the upright tower are represented in Fig. 12-8c.
The gravitational force , taken to act at the com, has a
vertical line of action and a moment arm of R(the per-
pendicular distance from the pivot to the line of action).
About the pivot, the torque associated with this force
would tend to create clockwise rotation and thus is nega-
tive. The normal force on the southern wall also has a
vertical line of action, and its moment arm is 2R. About
the pivot, the torque associated with this force would
tend to create counterclockwise rotation and thus is posi-
tive. We now can write the torque-balancing equation
(tnet,z0) as
(R)(mg)(2R)(FNR)0,
which yields
We should have been able to guess this result: With the
center of mass located on the central axis (the cylinder’s
line of symmetry), the right side supports half the cylin-
der’s weight.
In Fig.12-8b, the com is shifted rightward by distance
The only change in the balance of torques equation is that
the moment arm for the gravitational force is now Rd
and the normal force at the right has a new magnitude
(Fig. 12-8d).Thus, we write
(Rd)(mg)(2R)(F
NR )0,
F
NR
d1
2h tan u.
FNR 1
2mg.
FNR
:
mg
:
FNR
:
FNL
:
Figure 12-8 A cylinder modeling the Tower of Pisa: (a) upright and
(b) leaning, with the center of mass shifted rightward. The forces
and moment arms to find torques about a pivot at point Ofor
the cylinder (c) upright and (d) leaning.
Additional examples, video, and practice available at WileyPLUS
com
1__
2
com
Ground
O
O
h
h
(a)
(c)
d
θ
O
(b)
2R
R
FNR
mg
O
(d)
2R
R + d
FNR
‘
mg
Indeterminate Structures
For the problems of this chapter, we have only three independent equations at
our disposal, usually two balance of forces equations and one balance-of-torques
equation about a given rotation axis. Thus, if a problem has more than three
unknowns, we cannot solve it.
Consider an unsymmetrically loaded car. What are the forces—all different—
on the four tires? Again, we cannot find them because we have only three inde-
pendent equations. Similarly, we can solve an equilibrium problem for a table
with three legs but not for one with four legs. Problems like these, in which there
are more unknowns than equations, are called indeterminate.
Yet solutions to indeterminate problems exist in the real world. If you rest
the tires of the car on four platform scales, each scale will register a definite read-
ing, the sum of the readings being the weight of the car.What is eluding us in our
efforts to find the individual forces by solving equations?
The problem is that we have assumed—without making a great point of
it—that the bodies to which we apply the equations of static equilibrium are
perfectly rigid. By this we mean that they do not deform when forces are ap-
plied to them. Strictly, there are no such bodies. The tires of the car, for ex-
ample, deform easily under load until the car settles into a position of static
equilibrium.
We have all had experience with a wobbly restaurant table, which we usually
level by putting folded paper under one of the legs. If a big enough elephant sat
on such a table, however, you may be sure that if the table did not collapse, it
338 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
12-3 ELASTICITY
Learning Objectives
12.10 For shearing, apply the equation that relates stress to
strain and the shear modulus.
12.11 For hydraulic stress, apply the equation that relates
fluid pressure to strain and the bulk modulus.
●Three elastic moduli are used to describe the elastic behav-
ior (deformations) of objects as they respond to forces that act
on them. The strain (fractional change in length) is linearly re-
lated to the applied stress (force per unit area) by the proper
modulus, according to the general stress–strain relation
stress modulus strain.
●When an object is under tension or compression, the
stress–strain relation is written as
where L/Lis the tensile or compressive strain of the object,
Fis the magnitude of the applied force causing the strain,
Ais the cross-sectional area over which is applied (per-
pendicular to A), and Eis the Young’s modulus for the ob-
ject. The stress is F/A.
F
:
F
:
F
AEL
L,
●When an object is under a shearing stress, the stress–strain
relation is written as
where x/Lis the shearing strain of the object, xis the
displacement of one end of the object in the direction of the
applied force , and Gis the shear modulus of the object.
The stress is F/A.
●When an object undergoes hydraulic compression due to a
stress exerted by a surrounding fluid, the stress–strain relation
is written as
where pis the pressure (hydraulic stress) on the object due
to the fluid, V/V(the strain) is the absolute value of the frac-
tional change in the object’s volume due to that pressure, and
Bis the bulk modulus of the object.
pBV
V,
F
:
F
AGx
L,
After reading this module, you should be able to . . .
12.07 Explain what an indeterminate situation is.
12.08 For tension and compression, apply the equation that
relates stress to strain and Young’s modulus.
12.09 Distinguish between yield strength and ultimate strength.
Key Ideas
would deform just like the tires of a car. Its legs would all touch the floor,
the forces acting upward on the table legs would all assume definite (and differ-
ent) values as in Fig. 12-9, and the table would no longer wobble. Of course, we
(and the elephant) would be thrown out onto the street but, in principle, how do
we find the individual values of those forces acting on the legs in this or similar
situations where there is deformation?
To solve such indeterminate equilibrium problems, we must supplement
equilibrium equations with some knowledge of elasticity, the branch of physics
and engineering that describes how real bodies deform when forces are applied
to them.
339
12-3 ELASTICITY
Checkpoint 3
A horizontal uniform bar of weight 10 N is to hang from a ceiling by two wires that
exert upward forces and on the bar.The figure shows four arrangements for the
wires.Which arrangements, if any, are indeterminate (so that we cannot solve for nu-
merical values of and )?
F
:
2
F
:
1
F
:
2
F
:
1
10 N 10 N
10 N 10 N
dd
dd/2
(a)(b)
(c)(d)
F1F1
F1F2
F2F2
F2
F1
com
Fg
F1
F2
F3
F4
Figure 12-9 The table is an indeterminate
structure.The four forces on the table legs
differ from one another in magnitude and
cannot be found from the laws of static
equilibrium alone.
Elasticity
When a large number of atoms come together to form a metallic solid, such as an
iron nail, they settle into equilibrium positions in a three-dimensional lattice, a
repetitive arrangement in which each atom is a well-defined equilibrium distance
from its nearest neighbors. The atoms are held together by interatomic forces
that are modeled as tiny springs in Fig. 12-10. The lattice is remarkably rigid,
which is another way of saying that the “interatomic springs” are extremely stiff.
It is for this reason that we perceive many ordinary objects, such as metal ladders,
tables, and spoons, as perfectly rigid. Of course, some ordinary objects, such as gar-
den hoses or rubber gloves, do not strike us as rigid at all. The atoms that make up
these objects do not form a rigid lattice like that of Fig. 12-10 but are aligned in long,
flexible molecular chains, each chain being only loosely bound to its neighbors.
All real “rigid” bodies are to some extent elastic, which means that we can
change their dimensions slightly by pulling, pushing, twisting, or compressing
them. To get a feeling for the orders of magnitude involved, consider a vertical
steel rod 1 m long and 1 cm in diameter attached to a factory ceiling. If you hang
a subcompact car from the free end of such a rod, the rod will stretch but only by
about 0.5 mm, or 0.05%. Furthermore, the rod will return to its original length
when the car is removed.
If you hang two cars from the rod, the rod will be permanently stretched and
will not recover its original length when you remove the load. If you hang three
cars from the rod, the rod will break. Just before rupture, the elongation of the
Figure 12-10 The atoms of a metallic solid are
distributed on a repetitive three-dimensional
lattice.The springs represent interatomic
forces.
rod will be less than 0.2%. Although deformations of this size seem small, they
are important in engineering practice. (Whether a wing under load will stay on an
airplane is obviously important.)
Three Ways. Figure 12-11 shows three ways in which a solid might change
its dimensions when forces act on it. In Fig. 12-11a, a cylinder is stretched. In
Fig. 12-11b, a cylinder is deformed by a force perpendicular to its long axis,
much as we might deform a pack of cards or a book. In Fig. 12-11c, a solid ob-
ject placed in a fluid under high pressure is compressed uniformly on all sides.
What the three deformation types have in common is that a stress, or deform-
ing force per unit area, produces a strain, or unit deformation. In Fig. 12-11, ten-
sile stress (associated with stretching) is illustrated in (a), shearing stress in (b),
and hydraulic stress in (c).
The stresses and the strains take different forms in the three situations of
Fig. 12-11, but—over the range of engineering usefulness—stress and strain are
proportional to each other.The constant of proportionality is called a modulus of
elasticity, so that
stress modulus strain. (12-22)
In a standard test of tensile properties, the tensile stress on a test cylinder
(like that in Fig. 12-12) is slowly increased from zero to the point at which the
cylinder fractures, and the strain is carefully measured and plotted.The result is a
graph of stress versus strain like that in Fig. 12-13. For a substantial range of
applied stresses, the stress–strain relation is linear, and the specimen recovers its
original dimensions when the stress is removed; it is here that Eq. 12-22 applies. If
the stress is increased beyond the yield strength Syof the specimen, the specimen
becomes permanently deformed. If the stress continues to increase, the specimen
eventually ruptures, at a stress called the ultimate strength Su.
Tension and Compression
For simple tension or compression, the stress on the object is defined as F/A,
where Fis the magnitude of the force applied perpendicularly to an area Aon
the object. The strain, or unit deformation, is then the dimensionless quantity
L/L, the fractional (or sometimes percentage) change in a length of the speci-
men. If the specimen is a long rod and the stress does not exceed the yield
strength, then not only the entire rod but also every section of it experiences
the same strain when a given stress is applied. Because the strain is dimension-
less, the modulus in Eq. 12-22 has the same dimensions as the stress—namely,
force per unit area.
340 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
Figure 12-11 (a) A cylinder subject to tensile stress stretches by an amount L.(b) A cylinder
subject to shearing stress deforms by an amount x, somewhat like a pack of playing cards
would. (c) A solid sphere subject to uniform hydraulic stress from a fluid shrinks in volume
by an amount V.All the deformations shown are greatly exaggerated.
L
Figure 12-12 A test specimen used to deter-
mine a stress–strain curve such as that of
Fig.12-13.The change Lthat occurs in a
certain length Lis measured in a tensile
stress–strain test.
Figure 12-13 A stress–strain curve for a steel
test specimen such as that of Fig.12-12.
The specimen deforms permanently when
the stress is equal to the yield strength of
the specimen’s material.It ruptures when the
stress is equal to the ultimate strength of the
material.
Ultimate
strength
Yield
strength
Range of permanent
deformation
Rupture
Linear (elastic) range
Strain (ΔL/L)
0
Stress (F/A)
ΔV
V
(c)
(a)
L + ΔL
L
F
F
(b)
L
Δx
F
F
341
12-3 ELASTICITY
Figure 12-14 A strain gage of overall dimen-
sions 9.8 mm by 4.6 mm.The gage is fas-
tened with adhesive to the object whose
strain is to be measured; it experiences
the same strain as the object.The electri-
cal resistance of the gage varies with the
strain, permitting strains up to 3% to be
measured.
Courtesy Micro Measurements, a Division
of Vishay Precision Group, Raleigh, NC
Table 12-1 Some Elastic Properties of Selected Materials of Engineering Interest
Young’s Ultimate Yield
Density rModulus EStrength SuStrength Sy
Material (kg/m3) (109N/m2) (106N/m2) (106N/m2)
Steela7860 200 400 250
Aluminum 2710 70 110 95
Glass 2190 65 50b—
Concretec2320 30 40b—
Woodd525 13 50b—
Bone 1900 9b170b—
Polystyrene 1050 3 48 —
aStructural steel (ASTM-A36). bIn compression.
cHigh strength dDouglas fir.
The modulus for tensile and compressive stresses is called the Young’s modulus
and is represented in engineering practice by the symbol E. Equation 12-22 becomes
(12-23)
The strain L/Lin a specimen can often be measured conveniently with a strain
gage (Fig. 12-14), which can be attached directly to operating machinery with an
adhesive. Its electrical properties are dependent on the strain it undergoes.
Although the Young’s modulus for an object may be almost the same for tension
and compression,the object’s ultimate strength may well be different for the two types
of stress. Concrete, for example, is very strong in compression but is so weak in tension
that it is almost never used in that manner.Table 12-1 shows the Young’s modulus and
other elastic properties for some materials of engineering interest.
Shearing
In the case of shearing, the stress is also a force per unit area, but the force vector
lies in the plane of the area rather than perpendicular to it. The strain is the
dimensionless ratio x/L, with the quantities defined as shown in Fig. 12-11b.The
corresponding modulus, which is given the symbol Gin engineering practice, is
called the shear modulus. For shearing, Eq. 12-22 is written as
(12-24)
Shearing occurs in rotating shafts under load and in bone fractures due to bending.
Hydraulic Stress
In Fig. 12-11c, the stress is the fluid pressure pon the object, and, as you will see in
Chapter 14, pressure is a force per unit area. The strain is V/V, where Vis the
original volume of the specimen and Vis the absolute value of the change in vol-
ume.The corresponding modulus, with symbol B, is called the bulk modulus of the
material. The object is said to be under hydraulic compression, and the pressure
can be called the hydraulic stress. For this situation, we write Eq. 12-22 as
(12-25)
The bulk modulus is 2.2 109N/m2for water and 1.6 1011 N/m2for
steel. The pressure at the bottom of the Pacific Ocean, at its average depth
of about 4000 m, is 4.0 107N/m2. The fractional compression V/Vof a volume
of water due to this pressure is 1.8%; that for a steel object is only about 0.025%. In
general, solids—with their rigid atomic lattices—are less compressible than
liquids, in which the atoms or molecules are less tightly coupled to their neighbors.
pBV
V.
F
AGx
L.
F
AEL
L.
342 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
2.2 108N/m2. (Answer)
The yield strength for structural steel is 2.5 108N/m2,so
this rod is dangerously close to its yield strength.
We find the value of Young’s modulus for steel in
Table 12-1.Then from Eq. 12-23 we find the elongation:
8.9 104m0.89 mm. (Answer)
For the strain, we have
1.1 1030.11%. (Answer)
L
L8.9 104 m
0.81 m
L(F/A)L
E(2.2 108 N/m2)(0.81 m)
2.0 1011 N/m2
stress F
AF
pR26.2 104 N
(p)(9.5 103 m)2
Sample Problem 12.05 Stress and strain of elongated rod
One end of a steel rod of radius R9.5 mm and length
L81 cm is held in a vise. A force of magnitude
F62 kN is then applied perpendicularly to the end face
(uniformly across the area) at the other end, pulling di-
rectly away from the vise. What are the stress on the rod
and the elongation Land strain of the rod?
KEY IDEAS
(1) Because the force is perpendicular to the end face and
uniform, the stress is the ratio of the magnitude Fof the
force to the area A. The ratio is the left side of Eq. 12-23.
(2) The elongation Lis related to the stress and Young’s
modulus Eby Eq. 12-23 (F/AEL/L). (3) Strain is the
ratio of the elongation to the initial length L.
Calculations: To find the stress, we write
mation gives us
(12-27)
We cannot solve this equation because it has two unknowns,
F4and F3.
To get a second equation containing F4and F3, we can use
a vertical yaxis and then write the balance of vertical forces
(Fnet,y0) as
3F3F4Mg 0, (12-28)
where Mg is equal to the magnitude of the gravitational force
on the system. (Three legs have force on them.) To solve
the simultaneous equations 12-27 and 12-28 for, say, F3,we
first use Eq. 12-28 to find that F4Mg 3F3. Substituting
that into Eq. 12-27 then yields, after some algebra,
548 N 5.5 102N. (Answer)
From Eq. 12-28 we then find
F4Mg 3F3(290 kg)(9.8 m/s2)3(548 N)
1.2 kN. (Answer)
You can show that the three short legs are each compressed
by 0.42 mm and the single long leg by 0.92 mm.
(5.0 104 m)(104 m2)(1.3 1010 N/m2)
(4)(1.00 m)
(290 kg)(9.8 m/s2)
4
F3Mg
4dAE
4L
F
:
3
F4L
AE F3L
AE d.
Sample Problem 12.06 Balancing a wobbly table
A table has three legs that are 1.00 m in length and a fourth
leg that is longer by d0.50 mm, so that the table wobbles
slightly. A steel cylinder with mass M290 kg is placed
on the table (which has a mass much less than M) so that
all four legs are compressed but unbuckled and the table
is level but no longer wobbles. The legs are wooden cylin-
ders with cross-sectional area A1.0 cm2; Young’s mod-
ulus is E1.3 1010 N/m2.What are the magnitudes of the
forces on the legs from the floor?
KEY IDEAS
We take the table plus steel cylinder as our system.The situ-
ation is like that in Fig. 12-9, except now we have a steel
cylinder on the table. If the tabletop remains level, the legs
must be compressed in the following ways: Each of the short
legs must be compressed by the same amount (call it L3)
and thus by the same force of magnitude F3. The single long
leg must be compressed by a larger amount L4and thus by
a force with a larger magnitude F4. In other words, for a
level tabletop, we must have
L4L3d. (12-26)
From Eq. 12-23, we can relate a change in length to the
force causing the change with LFL/AE, where Lis the
original length of a leg.We can use this relation to replace L4
and L3in Eq. 12-26. However, note that we can approximate
the original length Las being the same for all four legs.
Calculations: Making those replacements and that approxi-
Additional examples, video, and practice available at WileyPLUS
343
QUESTIONS
Static Equilibrium A rigid body at rest is said to be in static
equilibrium. For such a body, the vector sum of the external forces
acting on it is zero:
(balance of forces). (12-3)
If all the forces lie in the xy plane, this vector equation is equiva-
lent to two component equations:
Fnet,x0 and Fnet,y0(balance of forces). (12-7, 12-8)
Static equilibrium also implies that the vector sum of the external
torques acting on the body about any point is zero, or
(balance of torques). (12-5)
If the forces lie in the xy plane, all torque vectors are parallel to the
zaxis, and Eq. 12-5 is equivalent to the single component equation
tnet,z0(balance of torques). (12-9)
Center of Gravity The gravitational force acts individually on
each element of a body.The net effect of all individual actions may
be found by imagining an equivalent total gravitational force
acting at the center of gravity. If the gravitational acceleration is
the same for all the elements of the body, the center of gravity is at
the center of mass.
Elastic Moduli Three elastic moduli are used to describe the
elastic behavior (deformations) of objects as they respond to
forces that act on them. The strain (fractional change in length) is
linearly related to the applied stress (force per unit area) by the
proper modulus, according to the general relation
stress modulus strain. (12-22)
g
:
F
:
g
t
:
net 0
F
:
net 0
Review & Summary
Tension and Compression When an object is under tension
or compression, Eq. 12-22 is written as
(12-23)
where L/Lis the tensile or compressive strain of the object, Fis
the magnitude of the applied force causing the strain, Ais the
cross-sectional area over which is applied (perpendicular to A,
as in Fig. 12-11a), and Eis the Young’s modulus for the object.The
stress is F/A.
Shearing When an object is under a shearing stress, Eq. 12-22 is
written as
(12-24)
where x/Lis the shearing strain of the object, xis the
displacement of one end of the object in the direction of the ap-
plied force (as in Fig. 12-11b), and Gis the shear modulus of the
object.The stress is F/A.
Hydraulic Stress When an object undergoes hydraulic com-
pression due to a stress exerted by a surrounding fluid, Eq. 12-22 is
written as
(12-25)
where pis the pressure (hydraulic stress) on the object due to the
fluid, V/V(the strain) is the absolute value of the fractional
change in the object’s volume due to that pressure, and Bis the
bulk modulus of the object.
pBV
V,
F
:
F
AGx
L,
F
:F
:
F
AEL
L,
Questions
1Figure 12-15 shows three situations in which the same
horizontal rod is supported by a hinge on a wall at one end and a
cord at its other end. Without written calculation, rank the situa-
tions according to the magnitudes of (a) the force on the rod
from the cord, (b) the vertical force on the rod from the hinge,
and (c) the horizontal force on the rod from the hinge, greatest
first.
compared to that of the safe.(a) Rank the positions according to
the force on post Adue to the safe, greatest compression first,
greatest tension last, and indicate where, if anywhere, the force is
zero. (b) Rank them according to the force on post B.
3Figure 12-17 shows four overhead views of rotating uniform
disks that are sliding across a frictionless floor. Three forces, of
magnitude F,2F, or 3F, act on each disk, either at the rim, at the
center,or halfway between rim and center.The force vectors rotate
along with the disks, and, in the “snapshots” of Fig. 12-17, point left
or right.Which disks are in equilibrium?
50° 50°
(1) (2) (3)
Figure 12-15 Question 1.
2In Fig.12-16, a rigid beam is at-
tached to two posts that are fas-
tened to a floor.A small but heavy
safe is placed at the six positions
indicated, in turn.Assume that the
mass of the beam is negligible
1 2 3 4 5 6
A B
Figure 12-16 Question 2.
(a) (b) (c) (d)
F
3F
2F
2F
F
F F
F
2F
2
F
F
F
Figure 12-17 Question 3.
4A ladder leans against a frictionless wall but is prevented from
falling because of friction between it and the ground. Suppose
you shift the base of the ladder toward the wall. Determine
whether the following become larger, smaller, or stay the same (in
magnitude): (a) the normal force on the ladder from the ground,
(b) the force on the ladder from the wall, (c) the static frictional
force on the ladder from the ground, and (d) the maximum value
fs,max of the static frictional force.
5Figure 12-18 shows a mobile of toy penguins hanging from a
ceiling. Each crossbar is horizontal, has negligible mass, and ex-
tends three times as far to the right of the wire supporting it as to
the left. Penguin 1 has mass m148 kg. What are the masses of
(a) penguin 2,(b) penguin 3, and (c) penguin 4?
344 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
What is the tension in the short cord
labeled with T?
9In Fig. 12-22, a vertical rod is
hinged at its lower end and attached
to a cable at its upper end.A horizon-
tal force is to be applied to the rod
as shown. If the point at which the
force is applied is moved up the rod,
does the tension in the cable increase,
decrease,or remain the same?
10 Figure 12-23 shows a horizon-
tal block that is suspended by two
wires, Aand B, which are identical
except for their original lengths. The
center of mass of the block is closer
to wire Bthan to wire A. (a)
Measuring torques about the
block’s center of mass, state whether
the magnitude of the torque due to wire Ais greater than, less
than, or equal to the magnitude of the torque due to wire B.(b)
Which wire exerts more force on the block? (c) If the wires are now
equal in length, which one was originally shorter (before the block
was suspended)?
11 The table gives the initial lengths of three rods and the
changes in their lengths when forces are applied to their ends to
put them under strain. Rank the rods according to their strain,
greatest first.
Initial Length Change in Length
Rod A2L0L0
Rod B4L02L0
Rod C10L04L0
12 A physical therapist gone wild has constructed the (station-
ary) assembly of massless pulleys and cords seen in Fig. 12-24.
One long cord wraps around all the pulleys, and shorter cords
suspend pulleys from the ceiling or weights from the pulleys.
Except for one, the weights (in newtons) are indicated. (a) What
is that last weight? (Hint: When a cord loops halfway around a
pulley as here, it pulls on the pulley with a net force that is twice
the tension in the cord.) (b) What is the tension in the short cord
labeled T?
F
:
a
12
3 4
6Figure 12-19 shows an overhead
view of a uniform stick on which
four forces act. Suppose we choose
a rotation axis through point O, cal-
culate the torques about that axis
due to the forces, and find that these
torques balance. Will the torques
balance if, instead, the rotation axis is chosen to be at (a) point A
(on the stick), (b) point B(on line with the stick), or (c) point C
(off to one side of the stick)? (d) Suppose, instead, that we find
that the torques about point Odo not balance. Is there another
point about which the torques will balance?
7In Fig. 12-20, a stationary 5 kg rod AC is held
against a wall by a rope and friction between rod
and wall.The uniform rod is 1 m long, and angle
. (a) If you are to find the magnitude of
the force on the rod from the rope with a sin-
gle equation,at what labeled point should a rota-
tion axis be placed? With that choice of axis and
counterclockwise torques positive, what is the
sign of (b) the torque due to the rod’s weight
and (c) the torque due to the pull on the rod
by the rope? (d) Is the magnitude of greater
than,less than,or equal to the magnitude of ?
8Three piñatas hang from the
(stationary) assembly of massless
pulleys and cords seen in Fig. 12-21.
One long cord runs from the ceiling
at the right to the lower pulley at
the left, looping halfway around all
the pulleys. Several shorter cords
suspend pulleys from the ceiling or
piñatas from the pulleys. The
weights (in newtons) of two piñatas
are given. (a) What is the weight of
the third piñata? (Hint: A cord that
loops halfway around a pulley pulls
on the pulley with a net force that
is twice the tension in the cord.) (b)
tw
tr
tr
tw
T
:
u30
O A
C
B
Figure 12-19 Question 6.
θ
B C A
D
Figure 12-20
Question 7.
Fa
Figure 12-22 Question 9.
com
A B
Figure 12-23 Question 10.
Figure 12-18 Question 5.
T
10
17
Figure 12-21 Question 8.
T
534
6
20
15
23
Figure 12-24 Question 12.
345
PROBLEMS
•7 A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long.
He places one end on the ground 2.5 m from a wall, rests the upper
end against a cracked window, and climbs the ladder. He is 3.0 m up
along the ladder when the window breaks. Neglect friction between
the ladder and window and assume that the base of the ladder does
not slip. When the window is on the verge of breaking, what are (a)
the magnitude of the force on the window from the ladder, (b) the
magnitude of the force on the ladder from the ground, and (c) the
angle (relative to the horizontal) of that force on the ladder?
•8 A physics Brady Bunch, whose weights in newtons are
indicated in Fig. 12-27, is balanced on a seesaw.What is the number
of the person who causes the largest torque about the rotation axis
at fulcrum f directed (a) out of the page and (b) into the page?
Module 12-1 Equilibrium
•1 Because gvaries so little over the extent of
most structures, any structure’s center of gravity
effectively coincides with its center of mass.
Here is a fictitious example where gvaries more
significantly. Figure 12-25 shows an array of
six particles, each with mass m, fixed to the edge
of a rigid structure of negligible mass. The dis-
tance between adjacent particles along the edge
is 2.00 m.The following table gives the value of g
(m/s2) at each particle’s location. Using the
coordinate system shown, find (a) the xcoordi-
nate xcom and (b) the ycoordinate ycom of the center of mass of the
six-particle system.Then find (c) the xcoordinate xcog and (d) the y
coordinate ycog of the center of gravity of the six-particle system.
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Particle gParticle g
1 8.00 4 7.40
2 7.80 5 7.60
3 7.60 6 7.80
Module 12-2 Some Examples of Static Equilibrium
•2 An automobile with a mass of 1360 kg has 3.05 m between the
front and rear axles. Its center of gravity is located 1.78 m behind
the front axle.With the automobile on level ground, determine the
magnitude of the force from the ground on (a) each front wheel
(assuming equal forces on the front wheels) and (b) each rear
wheel (assuming equal forces on the rear wheels).
•3 In Fig. 12-26, a uniform sphere
of mass m0.85 kg and radius r4.2 cm is
held in place by a massless rope attached to a
frictionless wall a distance L8.0 cm above
the center of the sphere. Find (a) the tension
in the rope and (b) the force on the sphere from
the wall.
•4 An archer’s bow is drawn at its midpoint
until the tension in the string is equal to the
force exerted by the archer. What is the angle
between the two halves of the string?
•5 A rope of negligible mass is stretched
horizontally between two supports that are 3.44 m
apart.When an object of weight 3160 N is hung at the center of the
rope, the rope is observed to sag by 35.0 cm.What is the tension in
the rope?
•6 A scaffold of mass 60 kg and length 5.0 m is supported in a
horizontal position by a vertical cable at each end. A window
washer of mass 80 kg stands at a point 1.5 m from one end. What is
the tension in (a) the nearer cable and (b) the farther cable?
ILW
WWWSSM
3 4
2 5
1 6
y
x
Figure 12-25
Problem 1.
L
r
Figure 12-26
Problem 3.
4 3 2 1 0 1 2 3 4 meters
220 330 440 560 560 440 330 220 newtons
1 2 3 4 5 6 7 8
f
Figure 12-27 Problem 8.
•9 A meter stick balances horizontally on a knife-edge at the
50.0 cm mark. With two 5.00 g coins stacked over the 12.0 cm
mark, the stick is found to balance at the 45.5 cm mark. What is the
mass of the meter stick?
•10 The system in Fig. 12-28 is in
equilibrium, with the string in the
center exactly horizontal. Block A
weighs 40 N, block Bweighs 50 N,
and angle fis 35°. Find (a) tension
T1, (b) tension T2, (c) tension T3, and
(d) angle u.
•11 Figure 12-29 shows a
diver of weight 580 N standing at the
end of a diving board with a length
of L4.5 m and negligible mass.
The board is fixed to two pedestals
(supports) that are separated by dis-
tance d1.5 m. Of the forces acting
on the board, what are the (a) magni-
tude and (b) direction (up or down)
of the force from the left pedestal
and the (c) magnitude and (d) direc-
tion (up or down) of the force from
the right pedestal? (e) Which
pedestal (left or right) is being
stretched, and (f) which pedestal is being compressed?
SSM
SSM
T2
T3
T1
θ φ
AB
Figure 12-28 Problem 10.
d
L
Figure 12-29 Problem 11.
•12 In Fig. 12-30, trying to get his car out of mud, a man ties one
end of a rope around the front bumper and the other end tightly
around a utility pole 18 m away. He then pushes sideways on the
rope at its midpoint with a force of 550 N, displacing the center of
the rope 0.30 m, but the car barely moves.What is the magnitude of
the force on the car from the rope? (The rope stretches somewhat.)
346 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
•17 In Fig. 12-34, a uniform beam of
weight 500 N and length 3.0 m is sus-
pended horizontally. On the left it is
hinged to a wall; on the right it is sup-
ported by a cable bolted to the wall at
distance Dabove the beam. The least
tension that will snap the cable is 1200
N. (a) What value of Dcorresponds to
that tension? (b) To prevent the cable
from snapping, should Dbe increased
or decreased from that value?
•18 In Fig. 12-35, horizontal scaf-
fold 2, with uniform mass m230.0
kg and length L22.00 m, hangs
from horizontal scaffold 1, with uni-
form mass m150.0 kg. A 20.0 kg
box of nails lies on scaffold 2, cen-
tered at distance d0.500 m from
the left end. What is the tension Tin
the cable indicated?
•19 To crack a certain nut in a nut-
cracker, forces with magnitudes of at
least 40 N must act on its shell from
both sides. For the nutcracker of Fig.
12-36, with distances L12 cm and
d2.6 cm, what are the force com-
ponents F(perpendicular to the
handles) corresponding to that 40 N?
•20 A bowler holds a bowling ball
(M7.2 kg) in the palm of his hand
(Fig. 12-37). His upper arm is vertical; his lower arm (1.8 kg) is
horizontal. What is the magnitude of (a) the force of the biceps
muscle on the lower arm and (b) the force between the bony
structures at the elbow contact point?
F
Figure 12-30 Problem 12.
•13 Figure 12-31 shows the
anatomical structures in the
lower leg and foot that are
involved in standing on tip-
toe, with the heel raised
slightly off the floor so that
the foot effectively contacts
the floor only at point P.
Assume distance a5.0 cm,
distance b15 cm, and the
person’s weight W900 N.
Of the forces acting on the
foot, what are the (a) magni-
tude and (b) direction (up or down) of the force at point Afrom
the calf muscle and the (c) magnitude and (d) direction (up or
down) of the force at point Bfrom the lower leg bones?
•14 In Fig. 12-32, a horizontal
scaffold, of length 2.00 m and uni-
form mass 50.0 kg, is suspended
from a building by two cables. The
scaffold has dozens of paint cans
stacked on it at various points. The total mass of the paint cans is
75.0 kg. The tension in the cable at the right is 722 N. How far
horizontally from that cable is the center of mass of the system of
paint cans?
•15 Forces , , and act on the structure of Fig. 12-33,
shown in an overhead view.We wish to put the structure in equilib-
rium by applying a fourth force, at a point such as P. The fourth
force has vector components and .We are given that a2.0 m,F
:
v
F
:
h
F
:
3
F
:
2
F
:
1
ILW
Figure 12-32 Problem 14.
a b P
A
B
Lower leg bones
Calf muscle
Figure 12-31 Problem 13.
b3.0 m, c1.0 m, F120 N, F210 N, and F35.0 N. Find (a)
Fh, (b) Fv,and (c) d.
y
x
O
d
b a
a
P
c
F1F2
F3
Fh
Fv
Figure 12-33 Problem 15.
•16 A uniform cubical crate is 0.750 m on each side and weighs
500 N. It rests on a floor with one edge against a very small, fixed
obstruction.At what least height above the floor must a horizontal
force of magnitude 350 N be applied to the crate to tip it?
Cable
Beam
D
Figure 12-34 Problem 17.
T = ?
1
2
d d d
L2
Figure 12-35 Problem 18.
L
d
F⊥
F⊥
Figure 12-36 Problem 19.
Biceps
Lower arm
(forearm plus
hand) center
of mass
Elbow
contact
point
M
33 cm
15 cm
4.0 cm
Figure 12-37 Problem 20.
••21 The system in Fig. 12-38 is
in equilibrium. A concrete block of
mass 225 kg hangs from the end of
the uniform strut of mass 45.0 kg. A
cable runs from the ground, over
the top of the strut, and down to the
block, holding the block in place.
For angles f30.0° and u45.0°,
find (a) the tension Tin the cable
and the (b) horizontal and (c) verti-
cal components of the force on the strut from the hinge.
ILW
θ
φ
T
Hinge
Strut
Figure 12-38 Problem 21.
••22 In Fig. 12-39, a 55 kg
rock climber is in a lie-back climb
along a fissure, with hands pulling on
one side of the fissure and feet
pressed against the opposite side.
The fissure has width w0.20 m,
and the center of mass of the climber
is a horizontal distance d0.40 m
from the fissure. The coefficient of
static friction between hands and
rock is m10.40, and between boots
and rock it is m21.2. (a) What is the
347
PROBLEMS
feet–ground contact point. If he is on the
verge of sliding, what is the coefficient of
static friction between feet and ground?
••27 In Fig. 12-44, a 15 kg block is
held in place via a pulley system. The
person’s upper arm is vertical; the fore-
arm is at angle u30° with the horizon-
tal. Forearm and hand together have a
mass of 2.0 kg, with a center of mass at dis-
tance d115 cm from the contact point of
the forearm bone and the upper-arm bone
(humerus). The triceps muscle pulls verti-
cally upward on the forearm at distance d22.5 cm behind that
contact point. Distance d3is 35 cm. What are the (a) magnitude and
(b) direction (up or down) of the force on the forearm from the tri-
ceps muscle and the (c) magnitude and (d) direction (up or down) of
the force on the forearm from the humerus?
h
w
com
d
Figure 12-39 Problem 22.
Figure 12-40 Problem 23.
θ
θ
Hinge
θ
φ
Figure 12-41 Problem 24.
If the climber encounters wet rock, so
that m1and m2are reduced, what hap-
pens to (c) the answer to (a) and (d)
the answer to (b)?
••23 In Fig. 12-40, one end of a
uniform beam of weight 222 N is
hinged to a wall; the other end is sup-
ported by a wire that makes angles
u30.0° with both wall and beam.
Find (a) the tension in the wire and the
(b) horizontal and (c) vertical compo-
nents of the force of the hinge on the
beam.
••24 In Fig. 12-41, a climber
with a weight of 533.8 N is held by a
belay rope connected to her climbing
harness and belay device; the force of
the rope on her has a line of action
through her center of mass. The indi-
cated angles are u40.0° and f
30.0°. If her feet are on the verge of
sliding on the vertical wall, what is the
coefficient of static friction between
her climbing shoes and the wall?
••25 In Fig. 12-42, what
WWWSSM
h
r
F
Figure 12-42 Problem 25.
x
y
a
d
com
L
Figure 12-43 Problem 26.
Block
com
Humerus
θ
d3
d1
d2
Triceps
Figure 12-44
Problem 27.
••28 In Fig. 12-45, suppose the
length Lof the uniform bar is 3.00 m
and its weight is 200 N. Also, let the
block’s weight W300 N and the an-
gle u30.0°. The wire can withstand
a maximum tension of 500 N. (a) What
is the maximum possible distance x
before the wire breaks? With the
block placed at this maximum x, what
are the (b) horizontal and (c) vertical
components of the force on the bar
from the hinge at A?
••29 A door has a height of 2.1 m
along a yaxis that extends vertically
upward and a width of 0.91 m along an
xaxis that extends outward from the
hinged edge of the door.A hinge 0.30 m
from the top and a hinge 0.30 m from
the bottom each support half the door’s
mass, which is 27 kg. In unit-vector
notation, what are the forces on the
com
L
x
C
B
θ
A
Figure 12-45
Problems 28 and 34.
door at (a) the top hinge and (b) the
bottom hinge?
••30 In Fig. 12-46, a 50.0 kg uniform
least horizontal pull by the hands and push by the feet that
will keep the climber stable? (b) For the horizontal pull of
(a), what must be the vertical distance hbetween hands and feet?
••26 In Fig. 12-43, a climber leans out against a vertical
ice wall that has negligible friction. Distance ais 0.914 m and dis-
tance Lis 2.10 m. His center of mass is distance d0.940 m from the
dvCable
Hinge
Rod
L
dh
L
Figure 12-46 Problem 30.
square sign, of edge length L2.00 m, is
hung from a horizontal rod of length
dh3.00 m and negligible mass. A ca-
ble is attached to the end of the rod
magnitude of (constant) force ap-
plied horizontally at the axle of the
wheel is necessary to raise the wheel
over a step obstacle of height
h3.00 cm? The wheel’s radius is r
6.00 cm, and its mass is m0.800 kg.
F
:
and to a point on the wall at distance dv4.00 m above the point
where the rod is hinged to the wall. (a) What is the tension in the
cable? What are the (b) magnitude and (c) direction (left or right)
of the horizontal component of the force on the rod from the wall,
and the (d) magnitude and (e) direction (up or down) of the verti-
cal component of this force?
••31 In Fig. 12-47, a
nonuniform bar is suspended
at rest in a horizontal position
by two massless cords. One
cord makes the angle u
36.9° with the vertical; the
other makes the angle f
53.1° with the vertical. If the
length Lof the bar is 6.10 m,
compute the distance xfrom
the left end of the bar to its center of mass.
••32 In Fig. 12-48, the driver of a car on a horizontal road makes
an emergency stop by applying the brakes so that all four wheels
lock and skid along the road. The coefficient of kinetic friction be-
tween tires and road is 0.40. The separation between the front and
rear axles is L4.2 m, and the center of mass of the car is located
at distance d1.8 m behind the front axle and distance h0.75 m
above the road. The car weighs 11 kN. Find the magnitude of (a)
the braking acceleration of the car, (b) the normal force on each
rear wheel, (c) the normal force on each front wheel, (d) the brak-
ing force on each rear wheel, and (e) the braking force on each
front wheel. (Hint: Although the car is not in translational equilib-
rium, it is in rotational equilibrium.)
348 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
the beam at distance yfrom the lower end. The beam remains
vertical because of a cable attached at the upper end, at angle u
with the horizontal. Figure 12-49bgives the tension Tin the cable
as a function of the position of the applied force given as a fraction
y/Lof the beam length.The scale of the Taxis is set by Ts600 N.
Figure 12-49cgives the magnitude Fhof the horizontal force on the
beam from the hinge, also as a function of y/L. Evaluate (a) angle u
and (b) the magnitude of .
••34 In Fig. 12-45, a thin horizontal bar AB of negligible weight
and length Lis hinged to a vertical wall at Aand supported at B
by a thin wire BC that makes an angle uwith the horizontal. A
block of weight Wcan be moved anywhere along the bar; its posi-
tion is defined by the distance xfrom the wall to its center of
mass. As a function of x, find (a) the tension in the wire, and the
(b) horizontal and (c) vertical components of the force on the bar
from the hinge at A.
••35 A cubical box is filled with sand and weighs 890
N. We wish to “roll” the box by pushing horizontally on one of the
upper edges. (a) What minimum force is required? (b) What mini-
mum coefficient of static friction between box and floor is re-
quired? (c) If there is a more efficient way to roll the box, find the
smallest possible force that would have to be applied directly to
the box to roll it. (Hint: At the onset of tipping,where is the normal
force located?)
••36 Figure 12-50 shows a 70 kg climber
hanging by only the crimp hold of one hand on
the edge of a shallow horizontal ledge in a
rock wall. (The fingers are pressed down to
gain purchase.) Her feet touch the rock wall
at distance H2.0 m directly below her
crimped fingers but do not provide any sup-
port. Her center of mass is distance a0.20
m from the wall. Assume that the force from
the ledge supporting her fingers is equally
shared by the four fingers. What are the values
of the (a) horizontal component Fhand (b)
vertical component Fvof the force on each
fingertip?
••37 In Fig. 12-51, a uniform plank, with a
length Lof 6.10 m and a weight of 445 N, rests
on the ground and against a frictionless roller at
the top of a wall of height h3.05 m.The plank
remains in equilibrium for any value of u70°
but slips if u70°. Find the coefficient of static
friction between the plank and the ground.
WWWSSM
F
:
a
Figure 12-47 Problem 31.
θφ
L
xcom
h
d
L
Figure 12-48
Problem 32.
••33 Figure 12-49ashows a vertical uniform beam of length L
that is hinged at its lower end. A horizontal force is applied toF
:
a
T (N)
Ts
0 0.2 0.4
y/L
0.6 0.8 1
Fh (N)
240
120
0 0.2 0.4
y/L
0.6 0.8 1
(b)(c)
(a)
Fa
θ
Cable
y
L
Figure 12-49 Problem 33.
y
x
H
a
Fh
Fv
com
Figure 12-50
Problem 36.
LRoller
h
θ
Figure 12-51 Problem 37.
••38 In Fig. 12-52, uniform beams A
and Bare attached to a wall with hinges
and loosely bolted together (there is
no torque of one on the other). Beam
Ahas length LA2.40 m and mass
54.0 kg; beam Bhas mass 68.0 kg.The
two hinge points are separated by dis-
tance d1.80 m. In unit-vector nota-
tion, what is the force on (a) beam A
due to its hinge, (b) beam Adue to
the bolt, (c) beam Bdue to its hinge,
and (d) beam Bdue to the bolt?
•••39 For the stepladder shown in
Fig. 12-53, sides AC and CE are each
2.44 m long and hinged at C. Bar BD
is a tie-rod 0.762 m long, halfway up.
A man weighing 854 N climbs 1.80 m
along the ladder. Assuming that the
floor is frictionless and neglecting the
mass of the ladder,find (a) the tension
in the tie-rod and the magnitudes of
the forces on the ladder from the floor
at (b) Aand (c) E.(Hint: Isolate parts
of the ladder in applying the equilib-
rium conditions.)
•••40 Figure 12-54ashows a horizon-
tal uniform beam of mass mband
length Lthat is supported on the left
by a hinge attached to a wall and on the right by a cable at angle uwith
the horizontal. A package of mass mpis positioned on the beam at a
distance xfrom the left end. The total mass is mbmp61.22 kg.
Figure 12-54bgives the tension Tin the cable as a function of the
package’s position given as a fraction x/Lof the beam length. The
scale of the Taxis is set by Ta 500 N and Tb700 N. Evaluate (a)
angle u,(b) mass mb,and (c) mass mp.
349
PROBLEMS
be 0.53. How far (in percent) up the ladder must the firefighter go
to put the ladder on the verge of sliding?
Module 12-3 Elasticity
•43 A horizontal aluminum rod 4.8 cm in diameter
projects 5.3 cm from a wall.A 1200 kg object is suspended from the
end of the rod.The shear modulus of aluminum is 3.0 1010 N/m2.
Neglecting the rod’s mass, find (a) the shear stress on the rod and
(b) the vertical deflection of the end of the rod.
•44 Figure 12-55 shows the
stress–strain curve for a material.
The scale of the stress axis is set by
s 300, in units of 106N/m2. What
are (a) the Young’s modulus and (b)
the approximate yield strength for
this material?
••45 In Fig. 12-56, a lead brick rests
horizontally on cylinders Aand B.
The areas of the top faces of the cylin-
ders are related by AA2AB; the
Young’s moduli of the cylinders
are related by EA2EB. The cylin-
ders had identical lengths before the
brick was placed on them. What
fraction of the brick’s mass is sup-
ported (a) by cylinder Aand (b) by
cylinder B? The horizontal distances
between the center of mass of the
brick and the centerlines of the
cylinders are dAfor cylinder Aand
dBfor cylinder B.(c) What is the ratio dA/dB?
••46 Figure 12-57 shows an approximate plot of stress ver-
sus strain for a spider-web thread, out to the point of breaking at a
strain of 2.00. The vertical axis scale is set by values a0.12
GN/m2,b0.30 GN/m2, and c0.80 GN/m2. Assume that the
thread has an initial length of 0.80 cm, an initial cross-sectional area of
8.0 1012 m2, and (during stretching) a constant volume. The
strain on the thread is the ratio of the change in the thread’s
length to that initial length, and the stress on the thread is the ra-
tio of the collision force to that initial cross-sectional area.
Assume that the work done on the thread by the collision force is
given by the area under the curve on the graph.Assume also that
when the single thread snares a flying insect, the insect’s kinetic
energy is transferred to the stretching of the thread. (a) How
much kinetic energy would put the thread on the verge of break-
ing? What is the kinetic energy of (b) a fruit fly of mass 6.00 mg
and speed 1.70 m/s and (c) a bumble bee of mass 0.388 g and
speed 0.420 m/s? Would (d) the fruit fly and (e) the bumble bee
break the thread?
ILWSSM
LA
y
A
B
x
Bolt
d
Figure 12-52 Problem 38.
B
A E
D
C
Figure 12-53 Problem 39.
Figure 12-54 Problem 40.
Cable
θ
L
x
(a)(b)
Tb
Ta00.20.4
x/L
0.6 0.8 1
T (N)
s
0
Stress (106 N/m2)
Strain
0 0.002 0.004
Figure 12-55 Problem 44.
A B
com of brick
dAdB
Figure 12-56 Problem 45.
Stress (GN/m2)
c
0
Strain
2.0
b
a
1.0 1.4
Figure 12-57 Problem 46.
•••41 A crate, in the form of a cube with edge lengths of 1.2 m, con-
tains a piece of machinery; the center of mass of the crate and its
contents is located 0.30 m above the crate’s geometrical center. The
crate rests on a ramp that makes an angle uwith the horizontal.As u
is increased from zero, an angle will be reached at which the crate
will either tip over or start to slide down the ramp. If the coefficient
of static friction msbetween ramp and crate is 0.60, (a) does the crate
tip or slide and (b) at what angle udoes this occur? If ms0.70,
(c) does the crate tip or slide and (d) at what angle udoes this occur?
(Hint: At the onset of tipping, where is the normal force located?)
•••42 In Fig. 12-7 and the associated sample problem, let the co-
efficient of static friction msbetween the ladder and the pavement
••47 A tunnel of length L150 m, height H7.2 m, and width
5.8 m (with a flat roof) is to be constructed at distance d60 m
beneath the ground. (See Fig. 12-58.) The tunnel roof is to be sup-
ported entirely by square steel columns, each with a cross-sectional
area of 960 cm2. The mass of 1.0 cm3of the ground material is 2.8 g.
(a) What is the total weight of the ground material the columns must
support? (b) How many columns are needed to keep the compres-
sive stress on each column at one-half its ultimate strength?
350 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
are forced against rigid walls at distances rA7.0 cm and rB4.0 cm
from the axle. Initially the stoppers touch the walls without being
compressed. Then force of magnitude 220 N is applied perpendi-
cular to the rod at a distance R5.0 cm from the axle. Find the mag-
nitude of the force compressing (a) stopper Aand (b) stopper B.
F
:
H
d
L
Figure 12-58 Problem 47.
••48 Figure 12-59 shows the
stress versus strain plot for an
aluminum wire that is stretched
by a machine pulling in opposite
directions at the two ends of the
wire. The scale of the stress axis is
set by s7.0, in units of
107N/m2. The wire has an initial
length of 0.800 m and an initial
cross-sectional area of 2.00 106
m2.How much work does the force
from the machine do on the wire to produce a strain of 1.00 103?
••49 In Fig. 12-60, a 103 kg uni-
form log hangs by two steel wires,
Aand B, both of radius 1.20 mm.
Initially, wire Awas 2.50 m long
and 2.00 mm shorter than wire B.
The log is now horizontal. What
are the magnitudes of the forces
on it from (a) wire Aand (b) wire
B? (c) What is the ratio dA/dB?
•••50 Figure 12-61 rep-
resents an insect caught at the mid-
point of a spider-web thread. The
thread breaks under a stress of
8.20 108N/m2and a strain of
2.00. Initially, it was horizontal
and had a length of 2.00 cm and a
cross-sectional area of 8.00
1012 m2. As the thread was stretched under the weight of the in-
sect, its volume remained constant. If the weight of the insect
puts the thread on the verge of breaking, what is the insect’s
mass? (A spider’s web is built to break if a potentially harmful in-
sect, such as a bumble bee, becomes snared in the web.)
•••51 Figure 12-62 is an overhead view of a rigid rod that turns
about a vertical axle until the identical rubber stoppers Aand B
Stress (107 N/m2)
s
0
Strain (10–3)
1.0
Figure 12-59 Problem 48.
W
ire AWire B
dAdB
com
Figure 12-60 Problem 49.
Figure 12-61 Problem 50.
Additional Problems
52 After a fall, a 95 kg rock climber finds himself dangling from
the end of a rope that had been 15 m long and 9.6 mm in diameter
but has stretched by 2.8 cm. For the rope, calculate (a) the strain,
(b) the stress, and (c) the Young’s modulus.
53 In Fig. 12-63, a rectangular
slab of slate rests on a bedrock sur-
face inclined at angle u26°. The
slab has length L43 m, thickness
T2.5 m, and width W12 m, and
1.0 cm3of it has a mass of 3.2 g. The
coefficient of static friction between
slab and bedrock is 0.39. (a)
Calculate the component of the
gravitational force on the slab parallel to the bedrock surface. (b)
Calculate the magnitude of the static frictional force on the slab.
By comparing (a) and (b), you can see that the slab is in danger of
sliding. This is prevented only by chance protrusions of bedrock.
(c) To stabilize the slab, bolts are to be driven perpendicular to the
bedrock surface (two bolts are shown). If each bolt has a cross-
sectional area of 6.4 cm2and will snap under a shearing stress of
3.6 108N/m2, what is the minimum number of bolts needed?
Assume that the bolts do not affect the normal force.
54 A uniform ladder whose length
is 5.0 m and whose weight is 400 N
leans against a frictionless vertical
wall. The coefficient of static friction
between the level ground and the
foot of the ladder is 0.46. What is the
greatest distance the foot of the lad-
der can be placed from the base of
the wall without the ladder immedi-
ately slipping?
55 In Fig. 12-64, block A
(mass 10 kg) is in equilibrium, but it
would slip if block B(mass 5.0 kg)
were any heavier. For angle u30°,
what is the coefficient of static fric-
tion between block Aand the sur-
face below it?
56 Figure 12-65ashows a uniform ramp between two buildings
that allows for motion between the buildings due to strong winds.
SSM
SSM
Stopper B
Stopper ArA
rB
R
Axle
F
Figure 12-62 Problem 51.
TL
θ
Figure 12-63 Problem 53.
A
B
θ
Figure 12-64 Problem 55.
At its left end, it is hinged to the building wall; at its right end, it has
a roller that can roll along the building wall. There is no vertical
force on the roller from the building, only a horizontal force with
magnitude Fh. The horizontal distance between the buildings is
D4.00 m. The rise of the ramp is h0.490 m. A man walks
across the ramp from the left. Figure 12-65bgives Fhas a function
of the horizontal distance xof the man from the building at the
left. The scale of the Fhaxis is set by a20 kN and b25 kN.
What are the masses of (a) the ramp and (b) the man?
351
PROBLEMS
60 In Fig. 12-69, a package of mass
mhangs from a short cord that is tied
to the wall via cord 1 and to the ceil-
ing via cord 2. Cord 1 is at angle f
40° with the horizontal; cord 2 is at an-
gle u. (a) For what value of uis the
tension in cord 2 minimized? (b) In
terms of mg, what is the minimum ten-
sion in cord 2?
61 The force in Fig. 12-70
keeps the 6.40 kg block and the pulleys in
equilibrium. The pulleys have negligible
mass and friction. Calculate the tension Tin
the upper cable. (Hint: When a cable wraps
halfway around a pulley as here, the magni-
tude of its net force on the pulley is twice
the tension in the cable.)
62 A mine elevator is supported by a sin-
gle steel cable 2.5 cm in diameter. The total
mass of the elevator cage and occupants is
670 kg. By how much does the cable stretch
when the elevator hangs by (a) 12 m of ca-
ble and (b) 362 m of cable? (Neglect the
mass of the cable.)
63 Four bricks of length L, identi-
cal and uniform, are stacked on top of one
another (Fig. 12-71) in such a way that
part of each extends beyond the
one beneath. Find, in terms of
L, the maximum values of
(a) a1, (b) a2, (c) a3, (d)
a4, and (e) h, such
that the stack is
in equilibrium,
on the verge of
falling.
F
:
ILW
D
x
h
a0 2
x (m)
4
b
Fh(kN)
(a) (b)
Figure 12-65 Problem 56.
57 In Fig. 12-66, a 10 kg sphere
is supported on a frictionless plane
inclined at angle u45° from the
horizontal. Angle fis 25°. Cal-
culate the tension in the cable.
58 In Fig.12-67a, a uniform 40.0 kg
beam is centered over two rollers.
Vertical lines across the beam mark
off equal lengths.Two of the lines are
centered over the rollers; a 10.0 kg
package of tamales is centered over
roller B.What are the magnitudes of
the forces on the beam from (a)
roller Aand (b) roller B? The beam
is then rolled to the left until the
right-hand end is centered over
roller B(Fig. 12-67b). What now are
the magnitudes of the forces on the
beam from (c) roller Aand (d)
roller B? Next, the beam is rolled to
the right. Assume that it has a
length of 0.800 m. (e) What hori-
zontal distance between the pack-
age and roller Bputs the beam on
the verge of losing contact with
roller A?
59 In Fig. 12-68, an 817 kg
construction bucket is suspended
by a cable Athat is attached at O
to two other cables Band C, mak-
ing angles u151.0° and u266.0°
with the horizontal. Find the ten-
sions in (a) cable A, (b) cable B,
and (c) cable C.(Hint: To avoid
solving two equations in two un-
knowns, position the axes as
shown in the figure.)
SSM
φ
θ
Cable
Figure 12-66 Problem 57.
A B
(a)
(b)
Figure 12-67 Problem 58.
θ φ
1 2
Figure 12-69 Problem 60.
x
O
C
B
θ
A
y
1
θ
2
Figure 12-68 Problem 59.
T
F
Figure 12-70
Problem 61.
64 In Fig. 12-72, two identical, uni-
form, and frictionless spheres, each of
mass m, rest in a rigid rectangular con-
tainer.A line connecting their centers is
at 45° to the horizontal. Find the
magnitudes of the forces on the spheres
from (a) the bottom of the container,
(b) the left side of the container,
(c) the right side of the container, and
(d) each other. (Hint: The force of one
sphere on the other is directed along
the center–center line.)
65 In Fig. 12-73, a uniform beam
with a weight of 60 N and a length of
3.2 m is hinged at its lower end, and
a horizontal force of magnitude
50 N acts at its upper end.The beam
is held vertical by a cable that makes
angle u25° with the ground and is
attached to the beam at height h
2.0 m. What are (a) the tension in
the cable and (b) the force on the
beam from the hinge in unit-vector
notation?
F
:
45°
Figure 12-72 Problem 64.
h
θ
F
x
y
Figure 12-73 Problem 65.
h
a4
a3
a2
a1
L
Figure 12-71
Problem 63.
66 A uniform beam is 5.0 m long
and has a mass of 53 kg. In Fig. 12-
74, the beam is supported in a hori-
zontal position by a hinge and a ca-
ble, with angle u60°. In unit-vector
notation,what is the force on the beam
from the hinge?
67 A solid copper cube has an edge
length of 85.5 cm. How much stress must be applied to the cube to
reduce the edge length to 85.0 cm? The bulk modulus of copper is
1.4 1011 N/m2.
68 A construction worker at-
tempts to lift a uniform beam off the
floor and raise it to a vertical posi-
tion. The beam is 2.50 m long and
weighs 500 N.At a certain instant the
worker holds the beam momentarily
at rest with one end at distance d
1.50 m above the floor, as shown in
Fig. 12-75, by exerting a force on
the beam, perpendicular to the
beam. (a) What is the magnitude P?
(b) What is the magnitude of the (net) force of the floor on the
beam? (c) What is the minimum value the coefficient of static
friction between beam and floor can have in order for the beam
not to slip at this instant?
69 In Fig. 12-76, a uniform rod of mass mis
hinged to a building at its lower end, while its upper
end is held in place by a rope attached to the wall. If
angle u160°, what value must angle u2have so
that the tension in the rope is equal to mg/2?
70 A 73 kg man stands on a level bridge of
length L. He is at distance L/4 from one end. The
bridge is uniform and weighs 2.7 kN. What are the
magnitudes of the vertical forces on the bridge from
its supports at (a) the end farther from him and (b) the nearer end?
71 A uniform cube of side length 8.0 cm rests on a horizon-
tal floor.The coefficient of static friction between cube and floor is
m. A horizontal pull is applied perpendicular to one of the verti-
cal faces of the cube, at a distance 7.0 cm above the floor on the
vertical midline of the cube face. The magnitude of is gradually
increased. During that increase, for what values of mwill the cube
eventually (a) begin to slide and (b) begin to tip? (Hint: At the
onset of tipping,where is the normal force located?)
72 The system in Fig. 12-77 is in equilibrium.The angles are u160°
and u220°, and the ball has mass M2.0 kg.What is the tension
in (a) string ab and (b) string bc?
P
:
P
:
SSM
SSM
P
:
352 CHAPTER 12 EQUILIBRIUM AND ELASTICITY
73 A uniform ladder is 10 m
long and weighs 200 N. In
Fig. 12-78, the ladder leans against
a vertical, frictionless wall at height
h8.0 m above the ground. A
horizontal force is applied to the
ladder at distance d2.0 m from
its base (measured along the lad-
der). (a) If force magnitude F50
N, what is the force of the ground
on the ladder, in unit-vector nota-
tion? (b) If F150 N, what is the
force of the ground on the ladder,
also in unit-vector notation? (c) Suppose the coefficient of static
friction between the ladder and the ground is 0.38; for what mini-
mum value of the force magnitude Fwill the base of the ladder
just barely start to move toward the wall?
74 A pan balance is made up of a rigid, massless rod with a hang-
ing pan attached at each end. The rod is supported at and free to
rotate about a point not at its center. It is balanced by unequal
masses placed in the two pans.When an unknown mass mis placed
in the left pan, it is balanced by a mass m1placed in the right pan;
when the mass mis placed in the right pan, it is balanced by a mass
m2in the left pan. Show that
75 The rigid square frame in
Fig. 12-79 consists of the four side bars
AB,BC,CD, and DA plus two diago-
nal bars AC and BD, which pass each
other freely at E. By means of the turn-
buckle G, bar AB is put under tension,
as if its ends were subject to horizontal,
outward forces of magnitude 535 N.
(a) Which of the other bars are in ten-
sion? What are the magnitudes of (b)
the forces causing the tension in those bars and (c) the forces caus-
ing compression in the other bars? (Hint: Symmetry considera-
tions can lead to considerable simplification in this problem.)
76 A gymnast with mass 46.0 kg
stands on the end of a uniform bal-
ance beam as shown in Fig. 12-80.The
beam is 5.00 m long and has a mass of
250 kg (excluding the mass of the two
supports). Each support is 0.540 m
from its end of the beam. In unit-vec-
tor notation, what are the forces on
the beam due to (a) support 1 and
(b) support 2?
77 Figure 12-81 shows a 300 kg
cylinder that is horizontal. Three
steel wires support the cylinder
from a ceiling.Wires 1 and 3 are at-
tached at the ends of the cylinder,
and wire 2 is attached at the cen-
ter. The wires each have a cross-
sectional area of 2.00 106m2.
Initially (before the cylinder was put in place) wires 1 and 3
were 2.0000 m long and wire 2 was 6.00 mm longer than that.
Now (with the cylinder in place) all three wires have been
stretched.What is the tension in (a) wire 1 and (b) wire 2?
T
:
m1m1m2.
F
:
SSM
θ
Cable
Beam
y
x
Figure 12-74 Problem 66.
d
P
Figure 12-75 Problem 68.
Figure 12-76
Problem 69.
Rod
Rope
1
θ
2
θ
θ
c
a
b
2
θ
1
M
Figure 12-77 Problem 72.
x
y
h
d
F
Figure 12-78 Problem 73.
T T
G
A B
D C
E
Figure 12-79 Problem 75.
x
y
1 2
Figure 12-80 Problem 76.
1 2 3
Ceiling
Figure 12-81 Problem 77.
78 In Fig. 12-82, a uniform beam of
length 12.0 m is supported by a hori-
zontal cable and a hinge at angle u
50.0°. The tension in the cable is 400
N. In unit-vector notation, what are
(a) the gravitational force on the
beam and (b) the force on the beam
from the hinge?
79 Four bricks of length L,
identical and uniform, are stacked
on a table in two ways, as shown in
Fig. 12-83 (compare with Problem
63). We seek to maximize the over-
hang distance hin both arrangements. Find the optimum distances
a1,a2,b1, and b2,and calculate hfor the two arrangements.
353
PROBLEMS
the loop with the rope hanging vertically when the child’s father
pulls on the child with a horizontal force and displaces the child
to one side. Just before the child is released from rest, the rope
makes an angle of 15with the vertical and the tension in the
rope is 280 N. (a) How much does the child weigh? (b) What is
the magnitude of the (horizontal) force of the father on the child
just before the child is released? (c) If the maximum horizontal
force the father can exert on the child is 93 N, what is the maxi-
mum angle with the vertical the rope can make while the father is
pulling horizontally?
85 Figure 12-85ashows details of a finger in the crimp hold
of the climber in Fig. 12-50. A tendon that runs from muscles in
the forearm is attached to the far bone in the finger. Along the
way, the tendon runs through several guiding sheaths called pul-
leys.The A2 pulley is attached to the first finger bone; the A4 pul-
ley is attached to the second finger bone.To pull the finger toward
the palm, the forearm muscles pull the tendon through the pul-
leys, much like strings on a marionette can be pulled to move parts
of the marionette. Figure 12-85bis a simplified diagram of the sec-
ond finger bone, which has length d. The tendon’s pull on the
bone acts at the point where the tendon enters the A4 pulley, at
distance d/3 along the bone. If the force components on each of
the four crimped fingers in Fig. 12-50 are Fh13.4 N and Fv
162.4 N, what is the magnitude of ? The result is probably tolera-
ble, but if the climber hangs by only one or two fingers, the A2 and
A4 pulleys can be ruptured, a common ailment among rock
climbers.
F
:
t
F
:
t
x
y
θ
Cable
Figure 12-82 Problem 78.
h
a2
a1
(a)
L
h
b2
b1
b1
b1
(b)
L
Figure 12-83 Problem 79.
y
x
A
B
θ
Figure 12-84 Problem 83.
86 A trap door in a ceiling is 0.91 m square, has a mass of 11 kg,
and is hinged along one side, with a catch at the opposite side. If
the center of gravity of the door is 10 cm toward the hinged side
from the door’s center, what are the magnitudes of the forces ex-
erted by the door on (a) the catch and (b) the hinge?
87 A particle is acted on by forces given, in newtons, by F
:
1
80 A cylindrical aluminum rod, with an initial length of 0.8000
m and radius 1000.0 mm, is clamped in place at one end and then
stretched by a machine pulling parallel to its length at its other
end. Assuming that the rod’s density (mass per unit volume)
does not change, find the force magnitude that is required of the
machine to decrease the radius to 999.9 mm. (The yield strength
is not exceeded.)
81 A beam of length Lis carried by three men, one man at one
end and the other two supporting the beam between them on a
crosspiece placed so that the load of the beam is equally divided
among the three men. How far from the beam’s free end is the
crosspiece placed? (Neglect the mass of the crosspiece.)
82 If the (square) beam in Fig. 12-6aand the associated sample
problem is of Douglas fir, what must be its thickness to keep the
compressive stress on it to of its ultimate strength?
83 Figure 12-84 shows a stationary arrangement of two crayon
boxes and three cords. Box Ahas a mass of 11.0 kg and is on a
ramp at angle u30.0°; box Bhas a mass of 7.00 kg and hangs on a
cord. The cord connected to box Ais parallel to the ramp, which is
frictionless. (a) What is the tension in the upper cord, and (b) what
angle does that cord make with the horizontal?
1
6
45°
80°
10°
Ft
Fh
Fv
Fc
d/3
(b)
Far
bone
Second
finger
bone
First
finger bone
A2 A4
Attached
Tendon
(a)
Figure 12-85 Problem 85.
84 A makeshift swing is constructed by making a loop in one end
of a rope and tying the other end to a tree limb.A child is sitting in
8.40 5.70 and 16.0 4.10 . (a) What are the xcomponent
and (b) ycomponent of the force that balances the sum of theseF
:
3
j
ˆ
iˆ
F
:
2
j
ˆ
iˆ
forces? (c) What angle does have relative to the xaxis?
88 The leaning Tower of Pisa is 59.1 m high and 7.44 m in diame-
ter.The top of the tower is displaced 4.01 m from the vertical.Treat
the tower as a uniform, circular cylinder. (a) What additional dis-
placement, measured at the top, would bring the tower to the verge
of toppling? (b) What angle would the tower then make with the
vertical?
F
:
3
What Is Physics?
One of the long-standing goals of physics is to understand the gravitational
force—the force that holds you to Earth, holds the Moon in orbit around Earth,
and holds Earth in orbit around the Sun. It also reaches out through the whole of
our Milky Way galaxy, holding together the billions and billions of stars in the
Galaxy and the countless molecules and dust particles between stars. We are
located somewhat near the edge of this disk-shaped collection of stars and other
matter, 2.6 104light-years (2.5 1020 m) from the galactic center, around
which we slowly revolve.
The gravitational force also reaches across intergalactic space, holding
together the Local Group of galaxies, which includes, in addition to the Milky
Way, the Andromeda Galaxy (Fig. 13-1) at a distance of 2.3 106light-years
away from Earth, plus several closer dwarf galaxies, such as the Large Magellanic
Cloud.The Local Group is part of the Local Supercluster of galaxies that is being
drawn by the gravitational force toward an exceptionally massive region of space
called the Great Attractor. This region appears to be about 3.0 108light-years
from Earth, on the opposite side of the Milky Way. And the gravitational force is
even more far-reaching because it attempts to hold together the entire universe,
which is expanding.
CHAPTER 13
Gravitation
13-1 NEWTON’S LAW OF GRAVITATION
After reading this module, you should be able to . . .
13.01 Apply Newton’s law of gravitation to relate the gravita-
tional force between two particles to their masses and
their separation.
13.02 Identify that a uniform spherical shell of matter attracts
a particle that is outside the shell as if all the shell’s mass
were concentrated as a particle at its center.
13.03 Draw a free-body diagram to indicate the gravitational
force on a particle due to another particle or a uniform,
spherical distribution of matter.
Key Ideas
Learning Objectives
354
●Any particle in the universe attracts any other particle with a
gravitational force whose magnitude is
(Newton’s law of gravitation),
where m1and m2are the masses of the particles, ris their
separation, and G(6.67 1011 Nm2/kg2)is the gravita-
tional constant.
FGm1m2
r2
●The gravitational force between extended bodies is found
by adding (integrating) the individual forces on individual par-
ticles within the bodies. However, if either of the bodies is a
uniform spherical shell or a spherically symmetric solid, the
net gravitational force it exerts on an external object may be
computed as if all the mass of the shell or body were located
at its center.
355
13-1 NEWTON’S LAW OF GRAVITATION
Figure 13-1 The Andromeda Galaxy.
Located 2.3 106light-years from us, and
faintly visible to the naked eye, it is very
similar to our home galaxy, the Milky Way.
Courtesy NASA
This force is also responsible for some of the most mysterious structures in
the universe: black holes. When a star considerably larger than our Sun burns
out, the gravitational force between all its particles can cause the star to collapse
in on itself and thereby to form a black hole.The gravitational force at the surface
of such a collapsed star is so strong that neither particles nor light can escape
from the surface (thus the term “black hole”). Any star coming too near a black
hole can be ripped apart by the strong gravitational force and pulled into the
hole. Enough captures like this yields a supermassive black hole. Such mysterious
monsters appear to be common in the universe. Indeed, such a monster lurks at
the center of our Milky Way galaxy—the black hole there, called Sagittarius A*,
has a mass of about 3.7 106solar masses.The gravitational force near this black
hole is so strong that it causes orbiting stars to whip around the black hole, com-
pleting an orbit in as little as 15.2 y.
Although the gravitational force is still not fully understood, the starting
point in our understanding of it lies in the law of gravitation of Isaac Newton.
Newton’s Law of Gravitation
Before we get to the equations, let’s just think for a moment about something
that we take for granted. We are held to the ground just about right, not so
strongly that we have to crawl to get to school (though an occasional exam may
leave you crawling home) and not so lightly that we bump our heads on the ceil-
ing when we take a step. It is also just about right so that we are held to the
ground but not to each other (that would be awkward in any classroom) or to the
objects around us (the phrase “catching a bus” would then take on a new mean-
ing). The attraction obviously depends on how much “stuff” there is in ourselves
and other objects: Earth has lots of “stuff” and produces a big attraction but an-
other person has less “stuff” and produces a smaller (even negligible) attraction.
Moreover, this “stuff” always attracts other “stuff,” never repelling it (or a hard
sneeze could put us into orbit).
In the past people obviously knew that they were being pulled downward
(especially if they tripped and fell over), but they figured that the downward
force was unique to Earth and unrelated to the apparent movement of astro-
nomical bodies across the sky. But in 1665, the 23-year-old Isaac Newton recog-
nized that this force is responsible for holding the Moon in its orbit. Indeed he
showed that every body in the universe attracts every other body.This tendency
of bodies to move toward one another is called gravitation, and the “stuff” that
is involved is the mass of each body. If the myth were true that a falling apple
inspired Newton to his law of gravitation, then the attraction is between the
mass of the apple and the mass of Earth. It is appreciable because the mass of
Earth is so large, but even then it is only about 0.8 N. The attraction between
two people standing near each other on a bus is (thankfully) much less (less than
1mN) and imperceptible.
The gravitational attraction between extended objects such as two people
can be difficult to calculate. Here we shall focus on Newton’s force law between
two particles (which have no size). Let the masses be m1and m2and rbe their
separation. Then the magnitude of the gravitational force acting on each due to
the presence of the other is given by
(Newton’s law of gravitation). (13-1)
Gis the gravitational constant:
G6.67 1011 Nm2/kg2
6.67 1011 m3/kgs2. (13-2)
FGm1m2
r2
356 CHAPTER 13 GRAVITATION
Figure 13-3 The apple pulls up on Earth
just as hard as Earth pulls down on the
apple.
F= 0.80 N
F= 0.80 N
A uniform spherical shell of matter attracts a particle that is outside the shell as if
all the shell’s mass were concentrated at its center.
Checkpoint 1
A particle is to be placed, in turn, outside four objects, each of mass m: (1) a large uni-
form solid sphere, (2) a large uniform spherical shell,(3) a small uniform solid sphere,
and (4) a small uniform shell. In each situation, the distance between the particle and
the center of the object is d. Rank the objects according to the magnitude of the gravi-
tational force they exert on the particle, greatest first.
Earth can be thought of as a nest of such shells, one within another and each shell at-
tracting a particle outside Earth’s surface as if the mass of that shell were at the cen-
ter of the shell.Thus, from the apple’s point of view, Earth does behave like a parti-
cle,one that is located at the center of Earth and has a mass equal to that of Earth.
Third-Law Force Pair. Suppose that, as in Fig. 13-3, Earth pulls down on an
apple with a force of magnitude 0.80 N. The apple must then pull up on Earth
with a force of magnitude 0.80 N, which we take to act at the center of Earth. In
the language of Chapter 5, these forces form a force pair in Newton’s third law.
Although they are matched in magnitude, they produce different accelerations
when the apple is released. The acceleration of the apple is about 9.8 m/s2, the fa-
miliar acceleration of a falling body near Earth’s surface. The acceleration of
Earth, however, measured in a reference frame attached to the center of mass of
the apple–Earth system, is only about 1 1025 m/s2.
In Fig. 13-2a, is the gravitational force acting on particle 1 (mass m1) due to
particle 2 (mass m2). The force is directed toward particle 2 and is said to be an
attractive force because particle 1 is attracted toward particle 2. The magnitude
of the force is given by Eq. 13-1.We can describe as being in the positive direc-
tion of an raxis extending radially from particle 1 through particle 2 (Fig. 13-2b).
We can also describe by using a radial unit vector r
ˆ(a dimensionless vector of
magnitude 1) that is directed away from particle 1 along the raxis (Fig. 13-2c).
From Eq. 13-1, the force on particle 1 is then
(13-3)
The gravitational force on particle 2 due to particle 1 has the same magnitude
as the force on particle 1 but the opposite direction. These two forces form a
third-law force pair, and we can speak of the gravitational force between the two
particles as having a magnitude given by Eq. 13-1. This force between two parti-
cles is not altered by other objects, even if they are located between the particles.
Put another way, no object can shield either particle from the gravitational force
due to the other particle.
The strength of the gravitational force—that is, how strongly two particles
with given masses at a given separation attract each other—depends on the
value of the gravitational constant G. If G—by some miracle—were suddenly
multiplied by a factor of 10, you would be crushed to the floor by Earth’s
attraction. If Gwere divided by this factor, Earth’s attraction would be so weak
that you could jump over a building.
Nonparticles. Although Newton’s law of gravitation applies strictly to particles,
we can also apply it to real objects as long as the sizes of the objects are small relative
to the distance between them. The Moon and Earth are far enough apart so that, to
a good approximation,we can treat them both as particles—but what about an apple
and Earth? From the point of view of the apple, the broad and level Earth, stretching
out to the horizon beneath the apple,certainly does not look like a particle.
Newton solved the apple–Earth problem with the shell theorem:
F
:Gm1m2
r2r
ˆ.
F
:
F
:
F
:
(b)
F
1
2r
Draw the vector with
its tail on particle 1
to show the pulling.
(c)
A unit vector points
along the radial axis.
r
1
2r
ˆ
r
F
1
2
(a)
This is the pull on
particle 1 due to
particle 2.
Figure 13-2 (a) The gravitational force
on particle 1 due to particle 2 is an at-
tractive force because particle 1 is at-
tracted to particle 2. (b) Force is di-
rected along a radial coordinate axis r
extending from particle 1 through par-
ticle 2. (c) is in the direction of a unit
vector r
ˆalong the raxis.
F
:
F
:
F
:
357
13-2 GRAVITATION AND THE PRINCIPLE OF SUPERPOSITION
13-2 GRAVITATION AND THE PRINCIPLE OF SUPERPOSITION
After reading this module, you should be able to . . .
13.04 If more than one gravitational force acts on a particle,
draw a free-body diagram showing those forces, with the
tails of the force vectors anchored on the particle.
13.05 If more than one gravitational force acts on a particle,
find the net force by adding the individual forces as
vectors.
●Gravitational forces obey the principle of superposition;
that is, if nparticles interact, the net force on a particle
labeled particle 1 is the sum of the forces on it from all the
other particles taken one at a time:
in which the sum is a vector sum of the forces on
particle 1 from particles 2, 3, ..., n.
F
:
1i
F
:
1,net
n
i2
F
:
1i,
F
:
1,net
●The gravitational force on a particle from an extended
body is found by first dividing the body into units of differential
mass dm, each of which produces a differential force on
the particle, and then integrating over all those units to find
the sum of those forces:
F
:
1dF
:
.
dF
:
F
:
1
Learning Objectives
Key Ideas
Gravitation and the Principle of Superposition
Given a group of particles, we find the net (or resultant) gravitational force on
any one of them from the others by using the principle of superposition. This is a
general principle that says a net effect is the sum of the individual effects. Here,
the principle means that we first compute the individual gravitational forces that
act on our selected particle due to each of the other particles.We then find the net
force by adding these forces vectorially, just as we have done when adding forces
in earlier chapters.
Let’s look at two important points in that last (probably quickly read) sen-
tence. (1) Forces are vectors and can be in different directions, and thus we must
add them as vectors, taking into account their directions. (If two people pull on
you in the opposite direction, their net force on you is clearly different than if
they pull in the same direction.) (2) We add the individual forces. Think how im-
possible the world would be if the net force depended on some multiplying factor
that varied from force to force depending on the situation, or if the presence of
one force somehow amplified the magnitude of another force. No, thankfully, the
world requires only simple vector addition of the forces.
For ninteracting particles, we can write the principle of superposition for the
gravitational forces on particle 1 as
1,net 12 13 14 15 1n. (13-4)
Here 1,net is the net force on particle 1 due to the other particles and, for exam-
ple, 13 is the force on particle 1 from particle 3. We can express this equation
more compactly as a vector sum:
(13-5)
Real Objects. What about the gravitational force on a particle from a real (ex-
tended) object? This force is found by dividing the object into parts small enough to
treat as particles and then using Eq. 13-5 to find the vector sum of the forces on the
particle from all the parts. In the limiting case, we can divide the extended object
into differential parts each of mass dm and each producing a differential force dF
:
F
:
1,net
n
i2
F
:
1i.
F
:
F
:
F
:
F
:
F
:
F
:
F
:
F
:
358 CHAPTER 13 GRAVITATION
Force 12 is directed in the positive direction of the yaxis (Fig.
13-4b) and has only the ycomponent F12. Similarly, 13 is di-
rected in the negative direction of the xaxis and has only the x
component F13 (Fig. 13-4c). (Note something important:We
draw the force diagrams with the tail of a force vector an-
chored on the particle experiencing the force. Drawing them
in other ways invites errors, especially on exams.)
To find the net force 1,net on particle 1, we must add
the two forces as vectors (Figs. 13-4dand e).We can do so on
a vector-capable calculator. However, here we note that
F13 and F12 are actually the xand ycomponents of 1,net.
Therefore, we can use Eq. 3-6 to find first the magnitude and
then the direction of 1,net.The magnitude is
4.1 106N. (Answer)
Relative to the positive direction of the xaxis, Eq. 3-6 gives
the direction of 1,net as
Is this a reasonable direction (Fig. 13-4f)? No, because the
direction of 1,net must be between the directions of 12 and
13. Recall from Chapter 3 that a calculator displays only
one of the two possible answers to a tan1function.We find
the other answer by adding 180°:
76° 180° 104°, (Answer)
which is a reasonable direction for 1,net (Fig. 13-4g).F
:
F
:F
:
F
:
tan1F12
F13
tan14.00 106N
1.00 106N76.
F
:
2(4.00 106 N)2(1.00 106 N)2
F
1,net 2(F
12)2(F
13)2
F
:
F
:
F
:
F
:
F
:
Sample Problem 13.01 Net gravitational force, 2D, three particles
Figure 13-4ashows an arrangement of three particles, parti-
cle 1 of mass m16.0 kg and particles 2 and 3 of mass m2
m34.0 kg, and distance a2.0 cm. What is the net gravi-
tational force 1,net on particle 1 due to the other particles?
KEY IDEAS
(1) Because we have particles, the magnitude of the gravita-
tional force on particle 1 due to either of the other particles is
given by Eq. 13-1 (FGm1m2/r2). (2) The direction of either
gravitational force on particle 1 is toward the particle responsi-
ble for it. (3) Because the forces are not along a single axis, we
cannot simply add or subtract their magnitudes or their compo-
nents to get the net force. Instead, we must add them as vectors.
Calculations: From Eq. 13-1, the magnitude of the force 12
on particle 1 from particle 2 is
(13-7)
4.00 106N.
Similarly, the magnitude of force 13 on particle 1 from
particle 3 is
(13-8)
1.00 106N.
(6.67 1011 m3/kgs2)(6.0 kg)(4.0 kg)
(0.040 m)2
F13 Gm1m3
(2a)2
F
:
(6.67 1011 m3/kgs2)(6.0 kg)(4.0 kg)
(0.020 m)2
F12 Gm1m2
a2
F
:
F
:
Additional examples, video, and practice available at WileyPLUS
on the particle.In this limit, the sum of Eq. 13-5 becomes an integral and we have
(13-6)
in which the integral is taken over the entire extended object and we drop the
subscript “net.” If the extended object is a uniform sphere or a spherical shell, we
can avoid the integration of Eq. 13-6 by assuming that the object’s mass is
concentrated at the object’s center and using Eq.13-1.
F
:
1dF
:
,
Checkpoint 2
The figure shows four arrangements of three particles
of equal masses. (a) Rank the arrangements according
to the magnitude of the net gravitational force on the
particle labeled m, greatest first. (b) In arrangement 2, is
the direction of the net force closer to the line of length
dor to the line of length D?
d
D
m
(1)
m
d
D
(2) (3) (4)
dD
m
mD
d
359
13-3 GRAVITATION NEAR EARTH’S SURFACE
A
(a)
m3
m2
2a
a
m1m1
x
y
(b)
xx
y
F12
m1
(c)
x
y
F13
m1
(d)
x
y
F1,net
F13
F12
m1
(e)
x
y
F1,net
F13
F12
(f)
x
yy
(g)
x
–76°
104°
This is the force
(pull) on particle 1
due to particle 2.
We want the forces
(pulls) on particle 1,
not the forces on
the other particles.
This is one way to
show the net force
on particle 1. Note
the head-to-tail
arrangement.
This is another way,
also a head-to-tail
arrangement.
A calculator's inverse
tangent can give this
for the angle.
But this is the
correct angle.
This is the force
(pull) on particle 1
due to particle 3.
Figure 13-4 (a) An arrangement of three particles.The force on particle 1 due to (b) particle 2
and (c) particle 3.(d)–(g) Ways to combine the forces to get the net force magnitude and orientation.
In WileyPLUS,this figure is available as an animation with voiceover.
13-3 GRAVITATION NEAR EARTH’S SURFACE
After reading this module, you should be able to . . .
13.06 Distinguish between the free-fall acceleration and the
gravitational acceleration.
13.07 Calculate the gravitational acceleration near but out-
side a uniform, spherical astronomical body.
13.08 Distinguish between measured weight and the magni-
tude of the gravitational force.
●The gravitational acceleration agof a particle (of mass m) is
due solely to the gravitational force acting on it. When the
particle is at distance rfrom the center of a uniform, spheri-
cal body of mass M, the magnitude Fof the gravitational
force on the particle is given by Eq. 13-1. Thus, by Newton’s
second law,
Fmag,
which gives
.
●Because Earth’s mass is not distributed uniformly, because
the planet is not perfectly spherical, and because it rotates,
the actual free-fall acceleration of a particle near Earth dif-
fers slightly from the gravitational acceleration , and the par-
ticle’s weight (equal to mg) differs from the magnitude of the
gravitational force on it.
a
:
g
g
:
agGM
r2
Learning Objectives
Key Ideas
360 CHAPTER 13 GRAVITATION
Figure 13-5 The density of Earth as a func-
tion of distance from the center.The limits
of the solid inner core, the largely liquid
outer core, and the solid mantle are shown,
but the crust of Earth is too thin to show
clearly on this plot.
Inner
core Outer
core
Mantle
Surface
Density (10
3
kg/m
3
)
14
12
10
8
6
4
2
00 1 2 3 4 5 6 7
Distance from center
(
106m
)
Gravitation Near Earth’s Surface
Let us assume that Earth is a uniform sphere of mass M. The magnitude of the
gravitational force from Earth on a particle of mass m, located outside Earth a
distance rfrom Earth’s center, is then given by Eq. 13-1 as
. (13-9)
If the particle is released, it will fall toward the center of Earth, as a result of the
gravitational force , with an acceleration we shall call the gravitational accelera-
tion g. Newton’s second law tells us that magnitudes Fand agare related by
Fmag. (13-10)
Now, substituting Ffrom Eq. 13-9 into Eq. 13-10 and solving for ag, we find
. (13-11)
Table 13-1 shows values of agcomputed for various altitudes above Earth’s
surface. Notice that agis significant even at 400 km.
Since Module 5-1, we have assumed that Earth is an inertial frame by ne-
glecting its rotation.This simplification has allowed us to assume that the free-fall
acceleration gof a particle is the same as the particle’s gravitational acceleration
(which we now call ag). Furthermore, we assumed that ghas the constant value
9.8 m/s2any place on Earth’s surface. However, any gvalue measured at a given
location will differ from the agvalue calculated with Eq. 13-11 for that location
for three reasons: (1) Earth’s mass is not distributed uniformly, (2) Earth is not a
perfect sphere, and (3) Earth rotates. Moreover, because gdiffers from ag, the
same three reasons mean that the measured weight mg of a particle differs from
the magnitude of the gravitational force on the particle as given by Eq. 13-9. Let
us now examine those reasons.
1. Earth’s mass is not uniformly distributed. The density (mass per unit volume)
of Earth varies radially as shown in Fig. 13-5, and the density of the crust
(outer section) varies from region to region over Earth’s surface.Thus, gvaries
from region to region over the surface.
2. Earth is not a sphere. Earth is approximately an ellipsoid, flattened at the
poles and bulging at the equator. Its equatorial radius (from its center point
out to the equator) is greater than its polar radius (from its center point out
to either north or south pole) by 21 km. Thus, a point at the poles is closer to
the dense core of Earth than is a point on the equator. This is one reason the
free-fall acceleration gincreases if you were to measure it while moving at sea
level from the equator toward the north or south pole. As you move, you are
actually getting closer to the center of Earth and thus, by Newton’s law of
gravitation, gincreases.
3. Earth is rotating. The rotation axis runs through the north and south poles of
Earth. An object located on Earth’s surface anywhere except at those poles
must rotate in a circle about the rotation axis and thus must have a centripetal
acceleration directed toward the center of the circle.This centripetal accelera-
tion requires a centripetal net force that is also directed toward that center.
To see how Earth’s rotation causes gto differ from ag, let us analyze a simple
situation in which a crate of mass mis on a scale at the equator. Figure 13-6a
shows this situation as viewed from a point in space above the north pole.
Figure 13-6b, a free-body diagram for the crate, shows the two forces on
the crate, both acting along a radial raxis that extends from Earth’s center.The
normal force on the crate from the scale is directed outward, in the positive
direction of the raxis. The gravitational force, represented with its equivalent
mg, is directed inward. Because it travels in a circle about the center of Earth
a
:
F
:
N
agGM
r2
a
:
F
:
FGMm
r2
Table 13-1 Variation of agwith Altitude
Altitude agAltitude
(km) (m/s2) Example
Mean Earth
0 9.83 surface
8.8 9.80 Mt. Everest
Highest crewed
36.6 9.71 balloon
Space shuttle
400 8.70 orbit
Communications
35 700 0.225 satellite
361
13-3 GRAVITATION NEAR EARTH’S SURFACE
as Earth turns, the crate has a centripetal acceleration directed toward
Earth’s center. From Eq. 10-23 (arv2r), we know this acceleration is equal to
v2R, where vis Earth’s angular speed and Ris the circle’s radius (approxi-
mately Earth’s radius). Thus, we can write Newton’s second law for forces
along the raxis (Fnet,rmar) as
FNmagm(v2R). (13-12)
The magnitude FNof the normal force is equal to the weight mg read on the scale.
With mg substituted for FN, Eq. 13-12 gives us
mg magm(v2R), (13-13)
which says
Thus, the measured weight is less than the magnitude of the gravitational force
on the crate, because of Earth’s rotation.
Acceleration Difference. To find a corresponding expression for gand ag,we
cancel mfrom Eq. 13-13 to write
gagv2R, (13-14)
which says
Thus, the measured free-fall acceleration is less than the gravitational accelera-
tion because of Earth’s rotation.
Equator. The difference between accelerations gand agis equal to v2Rand
is greatest on the equator (for one reason, the radius of the circle traveled by the
crate is greatest there). To find the difference, we can use Eq. 10-5 (vu/t)
and Earth’s radius R6.37 106m. For one rotation of Earth, uis 2prad and
the time period tis about 24 h. Using these values (and converting hours to sec-
onds), we find that gis less than agby only about 0.034 m/s2(small compared to
9.8 m/s2). Therefore, neglecting the difference in accelerations gand agis often
justified. Similarly, neglecting the difference between weight and the magnitude
of the gravitational force is also often justified.
free-fall
acceleration
gravitational
acceleration
centripetal
acceleration
.
measured
weight
magnitude of
gravitational force
mass times
centripetal acceleration
.
a
:
Figure 13-6 (a) A crate sitting on a scale at Earth’s equator,as seen by an observer
positioned on Earth’s rotation axis at some point above the north pole. (b) A free-body
diagram for the crate, with a radial raxis extending from Earth’s center.The gravitational
force on the crate is represented with its equivalent mg.The normal force on the crate
from the scale is . Because of Earth’s rotation,the crate has a centripetal acceleration
that is directed toward Earth’s center.
a
:
F
:
N
a
:
North
pole
R
Scale
Crate
(a)
Two forces act
on this crate.
mag
r
(b)
Crate
a
FN
The normal force
is upward.
The gravitational
force is downward.
The net
force is
toward
the center.
So, the
crate's
acceleration
is too.
362 CHAPTER 13 GRAVITATION
13-4 GRAVITATION INSIDE EARTH
Learning Objectives
13.10 Calculate the gravitational force that is exerted on a
particle at a given radius inside a nonrotating uniform
sphere of matter.
●A uniform shell of matter exerts no net gravitational force on
a particle located inside it.
●The gravitational force on a particle inside a uniform
solid sphere, at a distance rfrom the center, is due only to
mass Mins in an “inside sphere” with that radius r:
Mins 4
3pr3rM
R3r3,
F
:
where ris the solid sphere’s density, Ris its radius, and Mis
its mass. We can assign this inside mass to be that of a parti-
cle at the center of the solid sphere and then apply Newton’s
law of gravitation for particles. We find that the magnitude of
the force acting on mass mis
FGmM
R3r.
After reading this module, you should be able to . . .
13.09 Identify that a uniform shell of matter exerts no net
gravitational force on a particle located inside it.
Key Ideas
where the MEvalue is taken from Appendix C. This result
means that the gravitational acceleration of the astronaut’s
feet toward Earth is slightly greater than the gravitational
acceleration of her head toward Earth. This difference in
acceleration (often called a tidal effect) tends to stretch her
body, but the difference is so small that she would never even
sense the stretching,much less suffer pain from it.
(b) If the astronaut is now “feet down” at the same orbital
radius r6.77 106m about a black hole of mass Mh
1.99 1031 kg (10 times our Sun’s mass), what is the dif-
ference between the gravitational acceleration at her feet
and at her head? The black hole has a mathematical sur-
face (event horizon) of radius Rh2.95 104m. Nothing,
not even light, can escape from that surface or anywhere
inside it. Note that the astronaut is well outside the sur-
face (at r229Rh).
Calculations: We again have a differential change dr in r
between the astronaut’s feet and head, so we can again use
Eq. 13-16. However, now we substitute Mh1.99 1031 kg
for ME.We find
14.5 m/s2.(Answer)
This means that the gravitational acceleration of the astro-
naut’s feet toward the black hole is noticeably larger than
that of her head.The resulting tendency to stretch her body
would be bearable but quite painful. If she drifted closer
to the black hole, the stretching tendency would increase
drastically.
dag2(6.67 1011 m3/kgs2)(1.99 1031 kg)
(6.77 106m)3 (1.70 m)
Sample Problem 13.02 Difference in acceleration at head and feet
(a) An astronaut whose height his 1.70 m floats “feet down”
in an orbiting space shuttle at distance r6.77 106m away
from the center of Earth. What is the difference between the
gravitational acceleration at her feet and at her head?
KEY IDEAS
We can approximate Earth as a uniform sphere of mass ME.
Then, from Eq. 13-11, the gravitational acceleration at any dis-
tance rfrom the center of Earth is
(13-15)
We might simply apply this equation twice, first with r
6.77 106m for the location of the feet and then with
r6.77 106m1.70 m for the location of the head.
However, a calculator may give us the same value for agtwice,
and thus a difference of zero, because his so much smaller
than r. Here’s a more promising approach: Because we have
a differential change dr in rbetween the astronaut’s feet and
head, we should differentiate Eq. 13-15 with respect to r.
Calculations: The differentiation gives us
(13-16)
where dagis the differential change in the gravitational
acceleration due to the differential change dr in r. For the
astronaut, dr hand r6.77 106m. Substituting data
into Eq. 13-16, we find
4.37 106m/s2,(Answer)
dag2(6.67 1011 m3/kgs2)(5.98 1024 kg)
(6.77 106m)3 (1.70 m)
dag2GME
r3dr,
agGME
r2.
Additional examples, video, and practice available at WileyPLUS
363
13-4 GRAVITATION INSIDE EARTH
Gravitation Inside Earth
Newton’s shell theorem can also be applied to a situation in which a particle is
located inside a uniform shell, to show the following:
A uniform shell of matter exerts no net gravitational force on a particle located
inside it.
Caution: This statement does not mean that the gravitational forces on the par-
ticle from the various elements of the shell magically disappear. Rather, it
means that the sum of the force vectors on the particle from all the elements
is zero.
If Earth’s mass were uniformly distributed, the gravitational force acting
on a particle would be a maximum at Earth’s surface and would decrease as
the particle moved outward, away from the planet. If the particle were to move
inward, perhaps down a deep mine shaft, the gravitational force would change
for two reasons. (1) It would tend to increase because the particle would
be moving closer to the center of Earth. (2) It would tend to decrease because
the thickening shell of material lying outside the particle’s radial position
would not exert any net force on the particle.
To find an expression for the gravitational force inside a uniform Earth, let’s
use the plot in Pole to Pole, an early science fiction story by George Griffith.Three
explorers attempt to travel by capsule through a naturally formed (and, of course,
fictional) tunnel directly from the south pole to the north pole. Figure 13-7 shows
the capsule (mass m) when it has fallen to a distance rfrom Earth’s center.At that
moment, the net gravitational force on the capsule is due to the mass Mins inside
the sphere with radius r(the mass enclosed by the dashed outline), not the mass
in the outer spherical shell (outside the dashed outline). Moreover, we can assume
that the inside mass Mins is concentrated as a particle at Earth’s center. Thus, we
can write Eq. 13-1, for the magnitude of the gravitational force on the capsule, as
(13-17)
Because we assume a uniform density r, we can write this inside mass in
terms of Earth’s total mass Mand its radius R:
Solving for Mins we find
(13-18)
Substituting the second expression for Mins into Eq. 13-17 gives us the magnitude
of the gravitational force on the capsule as a function of the capsule’s distance r
from Earth’s center:
(13-19)
According to Griffith’s story, as the capsule approaches Earth’s center, the gravita-
tional force on the explorers becomes alarmingly large and, exactly at the center, it
suddenly but only momentarily disappears. From Eq. 13-19 we see that, in fact, the
force magnitude decreases linearly as the capsule approaches the center, until it is
zero at the center.At least Griffith got the zero-at-the-center detail correct.
FGmM
R3r.
Mins 4
3pr3rM
R3r3.
rMins
4
3pr3M
4
3pR3.
density inside mass
inside volume total mass
total volume ,
FGmMins
r2.
m
r
Mins
Figure 13-7 A capsule of mass mfalls from
rest through a tunnel that connects Earth’s
south and north poles.When the capsule is
at distance rfrom Earth’s center, the por-
tion of Earth’s mass that is contained in a
sphere of that radius is Mins.
364 CHAPTER 13 GRAVITATION
●The gravitational potential energy U(r)of a system of two
particles, with masses Mand mand separated by a distance
r, is the negative of the work that would be done by the gravi-
tational force of either particle acting on the other if the sepa-
ration between the particles were changed from infinite (very
large) to r. This energy is
(gravitational potential energy).
●If a system contains more than two particles, its total
gravitational potential energy Uis the sum of the terms rep-
UGMm
r
resenting the potential energies of all the pairs. As an
example, for three particles, of masses m1,m2, and m3,
●An object will escape the gravitational pull of an astronomi-
cal body of mass Mand radius R(that is, it will reach an infi-
nite distance) if the object’s speed near the body’s surface is
at least equal to the escape speed, given by
vA2GM
R.
U
Gm1m2
r12
Gm1m3
r13
Gm2m3
r23
.
Key Ideas
Equation 13-19 can also be written in terms of the force vector and the
capsule’s position vector along a radial axis extending from Earth’s center.
Letting Krepresent the collection of constants in Eq. 13-19, we can rewrite the
force in vector form as
K, (13-20)
in which we have inserted a minus sign to indicate that and have opposite
directions. Equation 13-20 has the form of Hooke’s law (Eq. 7-20, k).
Thus, under the idealized conditions of the story, the capsule would oscillate like a
block on a spring, with the center of the oscillation at Earth’s center.After the cap-
sule had fallen from the south pole to Earth’s center, it would travel from the center
to the north pole (as Griffith said) and then back again, repeating the cycle forever.
For the real Earth, which certainly has a nonuniform distribution of mass
(Fig. 13-5), the force on the capsule would initially increase as the capsule de-
scends. The force would then reach a maximum at a certain depth, and only then
would it begin to decrease as the capsule further descends.
d
:
F
:
r
:
F
:
r
:
F
:
r
:
F
:
13-5 GRAVITATIONAL POTENTIAL ENERGY
After reading this module, you should be able to . . .
13.11 Calculate the gravitational potential energy of a system
of particles (or uniform spheres that can be treated as
particles).
13.12 Identify that if a particle moves from an initial point to
a final point while experiencing a gravitational force, the
work done by that force (and thus the change in gravita-
tional potential energy) is independent of which path is
taken.
13.13 Using the gravitational force on a particle near an as-
tronomical body (or some second body that is fixed in
place), calculate the work done by the force when the
body moves.
13.14 Apply the conservation of mechanical energy (includ-
ing gravitational potential energy) to a particle moving rela-
tive to an astronomical body (or some second body that is
fixed in place).
13.15 Explain the energy requirements for a particle to es-
cape from an astronomical body (usually assumed to be a
uniform sphere).
13.16 Calculate the escape speed of a particle in leaving an
astronomical body.
Learning Objectives
Gravitational Potential Energy
In Module 8-1, we discussed the gravitational potential energy of a particle–
Earth system. We were careful to keep the particle near Earth’s surface, so that
we could regard the gravitational force as constant.We then chose some reference
configuration of the system as having a gravitational potential energy of zero.
Often, in this configuration the particle was on Earth’s surface. For particles not
365
13-5 GRAVITATIONAL POTENTIAL ENERGY
on Earth’s surface, the gravitational potential energy decreased when the separa-
tion between the particle and Earth decreased.
Here, we broaden our view and consider the gravitational potential energy U
of two particles, of masses mand M, separated by a distance r.We again choose a
reference configuration with Uequal to zero. However, to simplify the equations,
the separation distance rin the reference configuration is now large enough to be
approximated as infinite. As before, the gravitational potential energy decreases
when the separation decreases. Since U0 for r, the potential energy is neg-
ative for any finite separation and becomes progressively more negative as the
particles move closer together.
With these facts in mind and as we shall justify next, we take the gravitational
potential energy of the two-particle system to be
(gravitational potential energy). (13-21)
Note that U(r) approaches zero as rapproaches infinity and that for any finite
value of r, the value of U(r) is negative.
Language. The potential energy given by Eq. 13-21 is a property of the sys-
tem of two particles rather than of either particle alone. There is no way to di-
vide this energy and say that so much belongs to one particle and so much to
the other. However, if as is true for Earth (mass M) and a baseball
(mass m), we often speak of “the potential energy of the baseball.” We can get
away with this because, when a baseball moves in the vicinity of Earth, changes
in the potential energy of the baseball–Earth system appear almost entirely as
changes in the kinetic energy of the baseball, since changes in the kinetic en-
ergy of Earth are too small to be measured. Similarly, in Module 13-7 we shall
speak of “the potential energy of an artificial satellite” orbiting Earth, because
the satellite’s mass is so much smaller than Earth’s mass. When we speak of the
potential energy of bodies of comparable mass, however, we have to be careful
to treat them as a system.
Multiple Particles. If our system contains more than two particles, we con-
sider each pair of particles in turn, calculate the gravitational potential energy of
that pair with Eq. 13-21 as if the other particles were not there, and then alge-
braically sum the results.Applying Eq. 13-21 to each of the three pairs of Fig. 13-8,
for example, gives the potential energy of the system as
. (13-22)
Proof of Equation 13-21
Let us shoot a baseball directly away from Earth along the path in Fig. 13-9. We
want to find an expression for the gravitational potential energy Uof the ball at
point Palong its path, at radial distance Rfrom Earth’s center.To do so, we first
find the work Wdone on the ball by the gravitational force as the ball travels
from point Pto a great (infinite) distance from Earth. Because the gravitational
force (r) is a variable force (its magnitude depends on r), we must use the tech-
niques of Module 7-5 to find the work. In vector notation, we can write
(13-23)
The integral contains the scalar (or dot) product of the force (r) and the differential
displacement vector dalong the ball’s path.We can expand that product as
(r)dF(r)dr cos f, (13-24)
where fis the angle between the directions of (r) and d. When we substituter
:
F
:
r
:
F
:
r
:
F
:
W
R
F
:
(r)dr
:
.
F
:
U
Gm1m2
r12
Gm1m3
r13
Gm2m3
r23
Mm,
UGMm
r
Figure 13-8 A system consisting of three par-
ticles.The gravitational potential energy of
the system is the sum of the gravitational
potential energies of all three pairs of
particles.
m2
m1
m3
r12
r13 r23
This pair has
potential energy.
Here too.
Here too.
Figure 13-9 A baseball is shot directly away
from Earth, through point Pat radial dis-
tance Rfrom Earth’s center.The gravita-
tional force on the ball and a differential
displacement vector dare shown, both di-
rected along a radial raxis.
r
:
F
:
M
P
r
dr
R
F
Work is done
as the baseball
moves upward.
366 CHAPTER 13 GRAVITATION
180° for fand Eq.13-1 for F(r), Eq. 13-24 becomes
where Mis Earth’s mass and mis the mass of the ball.
Substituting this into Eq. 13-23 and integrating give us
(13-25)
where Wis the work required to move the ball from point P(at distance R) to
infinity. Equation 8-1 (UW) tells us that we can also write that work in terms
of potential energies as
UUW.
Because the potential energy Uat infinity is zero, Uis the potential energy at P,
and Wis given by Eq. 13-25, this equation becomes
Switching Rto rgives us Eq. 13-21, which we set out to prove.
Path Independence
In Fig. 13-10, we move a baseball from point Ato point Galong a path consisting
of three radial lengths and three circular arcs (centered on Earth). We are inter-
ested in the total work Wdone by Earth’s gravitational force on the ball as it
moves from Ato G. The work done along each circular arc is zero, because the
direction of is perpendicular to the arc at every point. Thus, Wis the sum of
only the works done by along the three radial lengths.
Now, suppose we mentally shrink the arcs to zero.We would then be moving
the ball directly from Ato Galong a single radial length. Does that change W?
No. Because no work was done along the arcs, eliminating them does not change
the work.The path taken from Ato Gnow is clearly different, but the work done
by is the same.
We discussed such a result in a general way in Module 8-1. Here is the point:
The gravitational force is a conservative force. Thus, the work done by the grav-
itational force on a particle moving from an initial point ito a final point fis
independent of the path taken between the points. From Eq. 8-1, the change U
in the gravitational potential energy from point ito point fis given by
UUfUiW. (13-26)
Since the work Wdone by a conservative force is independent of the actual path
taken, the change Uin gravitational potential energy is also independent of the
path taken.
Potential Energy and Force
In the proof of Eq. 13-21, we derived the potential energy function U(r) from the
force function (r).We should be able to go the other way—that is, to start from
the potential energy function and derive the force function. Guided by Eq. 8-22
(F(x)dU(x)/dx), we can write
(13-27)GMm
r2.
FdU
dr d
dr
GMm
r
F
:
F
:
F
:
F
:
F
:
UWGMm
R.
0GMm
RGMm
R,
WGMm
R
1
r2dr
GMm
r
R
F
:
(r)dr
:GMm
r2dr,
Figure 13-10 Near Earth, a baseball is moved
from point Ato point Galong a path
consisting of radial lengths and circular arcs.
F
E
G
D
A
C
B
Earth
Actual path
from A to G
is irrelevant.
367
13-5 GRAVITATIONAL POTENTIAL ENERGY
This is Newton’s law of gravitation (Eq. 13-1). The minus sign indicates that the
force on mass mpoints radially inward, toward mass M.
Escape Speed
If you fire a projectile upward, usually it will slow, stop momentarily, and return
to Earth. There is, however, a certain minimum initial speed that will cause it to
move upward forever, theoretically coming to rest only at infinity. This minimum
initial speed is called the (Earth) escape speed.
Consider a projectile of mass m, leaving the surface of a planet (or some
other astronomical body or system) with escape speed v. The projectile has a
kinetic energy Kgiven by mv2and a potential energy Ugiven by Eq. 13-21:
in which Mis the mass of the planet and Ris its radius.
When the projectile reaches infinity, it stops and thus has no kinetic energy.
It also has no potential energy because an infinite separation between two bod-
ies is our zero-potential-energy configuration. Its total energy at infinity is
therefore zero. From the principle of conservation of energy, its total energy at
the planet’s surface must also have been zero, and so
This yields (13-28)
Note that vdoes not depend on the direction in which a projectile is fired
from a planet. However, attaining that speed is easier if the projectile is fired in
the direction the launch site is moving as the planet rotates about its axis. For
example, rockets are launched eastward at Cape Canaveral to take advantage of
the Cape’s eastward speed of 1500 km/h due to Earth’s rotation.
Equation 13-28 can be applied to find the escape speed of a projectile from
any astronomical body, provided we substitute the mass of the body for Mand
the radius of the body for R.Table 13-2 shows some escape speeds.
vA2GM
R.
KU1
2mv2
GMm
R
0.
UGMm
R,
1
2
Table 13-2 Some Escape Speeds
Body Mass (kg) Radius (m) Escape Speed (km/s)
Ceresa1.17 1021 3.8 1050.64
Earth’s moona7.36 1022 1.74 1062.38
Earth 5.98 1024 6.37 10611.2
Jupiter 1.90 1027 7.15 10759.5
Sun 1.99 1030 6.96 108618
Sirius Bb21030 11075200
Neutron starc21030 11042105
aThe most massive of the asteroids.
bAwhite dwarf (a star in a final stage of evolution) that is a companion of the bright star Sirius.
cThe collapsed core of a star that remains after that star has exploded in a supernova event.
Checkpoint 3
You move a ball of mass maway from a sphere of mass M. (a) Does the gravitational
potential energy of the system of ball and sphere increase or decrease? (b) Is positive
work or negative work done by the gravitational force between the ball and the sphere?
368 CHAPTER 13 GRAVITATION
13-6 PLANETS AND SATELLITES: KEPLER’S LAWS
After reading this module, you should be able to . . .
13.17 Identify Kepler’s three laws.
13.18 Identify which of Kepler’s laws is equivalent to the law
of conservation of angular momentum.
13.19 On a sketch of an elliptical orbit, identify the semimajor
axis, the eccentricity, the perihelion, the aphelion, and the
focal points.
13.20 For an elliptical orbit, apply the relationships between
the semimajor axis, the eccentricity, the perihelion, and the
aphelion.
13.21 For an orbiting natural or artificial satellite, apply
Kepler’s relationship between the orbital period and radius
and the mass of the astronomical body being orbited.
●The motion of satellites, both natural and artificial, is gov-
erned by Kepler’s laws:
1. The law of orbits. All planets move in elliptical orbits with
the Sun at one focus.
2. The law of areas. A line joining any planet to the Sun
sweeps out equal areas in equal time intervals. (This
statement is equivalent to conservation of angular
momentum.)
3. The law of periods. The square of the period Tof any
planet is proportional to the cube of the semimajor axis a
of its orbit. For circular orbits with radius r,
(law of periods),
where Mis the mass of the attracting body—the Sun in the
case of the solar system. For elliptical planetary orbits, the
semimajor axis ais substituted for r.
T2
4
2
GM
r3
Learning Objectives
Key Ideas
tially at distance 10REand finally at distance RE, where REis
Earth’s radius (6.37 106m). Substituting Eq. 13-21 for U
and for K, we rewrite Eq. 13-29 as
Rearranging and substituting known values, we find
(12 103m/s)2
2.567 108m2/s2,
and vf1.60 104m/s 16 km/s. (Answer)
At this speed, the asteroid would not have to be par-
ticularly large to do considerable damage at impact. If it
were only 5 m across, the impact could release about as
much energy as the nuclear explosion at Hiroshima.
Alarmingly, about 500 million asteroids of this size are
near Earth’s orbit, and in 1994 one of them apparently
penetrated Earth’s atmosphere and exploded 20 km
above the South Pacific (setting off nuclear-explosion
warnings on six military satellites).
2(6.67 1011 m3/kgs2)(5.98 1024 kg)
6.37 106 m 0.9
v2
fv2
i2GM
RE
11
10
1
2mv2
fGMm
RE
1
2mv2
iGMm
10RE
.
1
2mv2
Sample Problem 13.03 Asteroid falling from space, mechanical energy
An asteroid, headed directly toward Earth, has a speed of
12 km/s relative to the planet when the asteroid is 10 Earth
radii from Earth’s center. Neglecting the effects of Earth’s
atmosphere on the asteroid, find the asteroid’s speed vf
when it reaches Earth’s surface.
KEY IDEAS
Because we are to neglect the effects of the atmosphere on
the asteroid, the mechanical energy of the asteroid–Earth
system is conserved during the fall. Thus, the final mechani-
cal energy (when the asteroid reaches Earth’s surface) is
equal to the initial mechanical energy.With kinetic energy K
and gravitational potential energy U, we can write this as
KfUfKiUi. (13-29)
Also, if we assume the system is isolated, the system’s lin-
ear momentum must be conserved during the fall. Therefore,
the momentum change of the asteroid and that of Earth must
be equal in magnitude and opposite in sign. However, because
Earth’s mass is so much greater than the asteroid’s mass, the
change in Earth’s speed is negligible relative to the change in
the asteroid’s speed. So, the change in Earth’s kinetic energy is
also negligible.Thus, we can assume that the kinetic energies in
Eq.13-29 are those of the asteroid alone.
Calculations: Let mrepresent the asteroid’s mass and M
represent Earth’s mass (5.98 1024 kg). The asteroid is ini-
Additional examples, video, and practice available at WileyPLUS
369
13-6 PLANETS AND SATELLITES: KEPLER’S LAWS
Planets and Satellites: Kepler’s Laws
The motions of the planets, as they seemingly wander against the background of
the stars, have been a puzzle since the dawn of history. The “loop-the-loop”
motion of Mars, shown in Fig. 13-11, was particularly baffling. Johannes Kepler
(1571–1630), after a lifetime of study, worked out the empirical laws that govern
these motions. Tycho Brahe (1546–1601), the last of the great astronomers to
make observations without the help of a telescope, compiled the extensive data
from which Kepler was able to derive the three laws of planetary motion that
now bear Kepler’s name. Later, Newton (1642–1727) showed that his law of
gravitation leads to Kepler’s laws.
In this section we discuss each of Kepler’s three laws. Although here we
apply the laws to planets orbiting the Sun, they hold equally well for satellites,
either natural or artificial, orbiting Earth or any other massive central body.
1. THE LAW OF ORBITS: All planets move in elliptical orbits,with the Sun at one focus.
Figure 13-12 shows a planet of mass mmoving in such an orbit around the Sun,
whose mass is M. We assume that so that the center of mass of the
planet–Sun system is approximately at the center of the Sun.
The orbit in Fig. 13-12 is described by giving its semimajor axis aand its
eccentricity e, the latter defined so that ea is the distance from the center of the
ellipse to either focus For F9.An eccentricity of zero corresponds to a circle, in
which the two foci merge to a single central point.The eccentricities of the plane-
tary orbits are not large; so if the orbits are drawn to scale, they look circular.The
eccentricity of the ellipse of Fig. 13-12, which has been exaggerated for clarity, is
0.74.The eccentricity of Earth’s orbit is only 0.0167.
Mm,
Figure 13-11 The path seen from Earth for the
planet Mars as it moved against a back-
ground of the constellation Capricorn during
1971.The planet’s position on four days is
marked.Both Mars and Earth are moving in
orbits around the Sun so that we see the
position of Mars relative to us;this relative
motion sometimes results in an apparent
loop in the path of Mars.
July 26 September 4
June 6
October 14
2. THE LAW OF AREAS: A line that connects a planet to the Sun sweeps out
equal areas in the plane of the planet’s orbit in equal time intervals; that is, the
rate dA/dt at which it sweeps out area Ais constant.
Figure 13-12 A planet of mass mmoving in
an elliptical orbit around the Sun.The Sun,
of mass M, is at one focus Fof the ellipse.
The other focus is F9, which is located in
empty space. The semimajor axis aof the
ellipse, the perihelion (nearest the Sun)
distance Rp, and the aphelion (farthest
from the Sun) distance Raare also shown.
Ra
Rp
M
m
a
ea ea
F F'
θ
r
The Sun is at
one of the two
focal points.
Qualitatively, this second law tells us that the planet will move most slowly when
it is farthest from the Sun and most rapidly when it is nearest to the Sun. As it
turns out, Kepler’s second law is totally equivalent to the law of conservation of
angular momentum. Let us prove it.
The area of the shaded wedge in Fig. 13-13aclosely approximates the area
swept out in time tby a line connecting the Sun and the planet, which are
separated by distance r. The area Aof the wedge is approximately the area of
Figure 13-13 (a) In time t, the line rconnecting the planet to the Sun moves through an
angle u,sweeping out an area A(shaded). (b) The linear momentum of the planet
and the components of .p
:
p
:
M
θ
θ
Δ ΔA
r
θ
rΔ
M
θ
r
p⊥
pr
(a)(b)
SunSun
p
The planet sweeps out
this area.
These are the
two momentum
components.
370 CHAPTER 13 GRAVITATION
Table 13-3 Kepler’s Law of Periods for the
Solar System
Semimajor T2/a3
Axis Period (1034
Planet a(1010 m) T(y) y2/m3)
Mercury 5.79 0.241 2.99
Venus 10.8 0.615 3.00
Earth 15.0 1.00 2.96
Mars 22.8 1.88 2.98
Jupiter 77.8 11.9 3.01
Saturn 143 29.5 2.98
Uranus 287 84.0 2.98
Neptune 450 165 2.99
Pluto 590 248 2.99
Checkpoint 4
Satellite 1 is in a certain circular orbit around a planet, while satellite 2 is in a larger
circular orbit.Which satellite has (a) the longer period and (b) the greater speed?
a triangle with base ruand height r. Since the area of a triangle is one-half of the
base times the height, Ar2u. This expression for Abecomes more exact
1
2
3. THE LAW OF PERIODS: The square of the period of any planet is propor-
tional to the cube of the semimajor axis of its orbit.
To see this, consider the circular orbit of Fig. 13-14, with radius r(the radius of
a circle is equivalent to the semimajor axis of an ellipse). Applying Newton’s
second law (Fma) to the orbiting planet in Fig.13-14 yields
(13-33)
Here we have substituted from Eq. 13-1 for the force magnitude Fand used
Eq. 10-23 to substitute v2rfor the centripetal acceleration. If we now use Eq. 10-20
to replace vwith 2p/T, where Tis the period of the motion, we obtain Kepler’s
third law:
(law of periods). (13-34)
The quantity in parentheses is a constant that depends only on the mass Mof the
central body about which the planet orbits.
Equation 13-34 holds also for elliptical orbits, provided we replace rwith
a, the semimajor axis of the ellipse.This law predicts that the ratio T2/a3has es-
sentially the same value for every planetary orbit around a given massive
body. Table 13-3 shows how well it holds for the orbits of the planets of the
solar system.
T2
4p2
GM
r3
GMm
r2(m)(v2r).
Figure 13-14 A planet of mass mmoving
around the Sun in a circular orbit of radius r.
θ
M
r
m
as t(hence u) approaches zero. The instantaneous rate at which area is being
swept out is then
(13-30)
in which vis the angular speed of the line connecting Sun and planet, as the line
rotates around the Sun.
Figure 13-13bshows the linear momentum of the planet, along with the radial
and perpendicular components of . From Eq. 11-20 the magnitude of
the angular momentum of the planet about the Sun is given by the product of r
and the component of perpendicular to r.Here, for a planet of mass m,
(13-31)
where we have replaced with its equivalent vr(Eq. 10-18). Eliminating r2v
between Eqs. 13-30 and 13-31 leads to
(13-32)
If dA/dt is constant, as Kepler said it is, then Eq. 13-32 means that Lmust also be
constant—angular momentum is conserved. Kepler’s second law is indeed
equivalent to the law of conservation of angular momentum.
dA
dt L
2m.
v
mr2v,
Lrp(r)(mv)(r)(mvr)
p
:
p,
L
:(Lrp),
p
:
p
:
dA
dt 1
2r2du
dt 1
2r2v,
371
13-7 SATELLITES: ORBITS AND ENERGY
Ra2aRp
(2)(2.7 1012 m) 8.9 1010 m
5.3 1012 m. (Answer)
Table 13-3 shows that this is a little less than the semimajor
axis of the orbit of Pluto. Thus, the comet does not get far-
ther from the Sun than Pluto.
(b) What is the eccentricity eof the orbit of comet Halley?
KEY IDEA
We can relate e,a, and Rpvia Fig. 13-12, in which we see that
ea aRp.
Calculation: We have
(13-36)
(Answer)
This tells us that, with an eccentricity approaching unity, this
orbit must be a long thin ellipse.
18.9 1010 m
2.7 1012 m 0.97.
eaRp
a1Rp
a
Sample Problem 13.04 Kepler’s law of periods, Comet Halley
Comet Halley orbits the Sun with a period of 76 years and, in
1986, had a distance of closest approach to the Sun, its peri-
helion distance Rp, of 8.9 1010 m.Table 13-3 shows that this
is between the orbits of Mercury and Venus.
(a) What is the comet’s farthest distance from the Sun,
which is called its aphelion distance Ra?
KEY IDEAS
From Fig.13-12, we see that RaRp2a, where ais the semi-
major axis of the orbit. Thus, we can find Raif we first find a.
We can relate ato the given period via the law of periods
(Eq. 13-34) if we simply substitute the semimajor axis afor r.
Calculations: Making that substitution and then solving for
a, we have
. (13-35)
If we substitute the mass Mof the Sun, 1.99 1030 kg, and
the period Tof the comet, 76 years or 2.4 109s, into
Eq. 13-35, we find that a2.7 1012 m. Now we have
a
GMT 2
4p2
1/3
Additional examples, video, and practice available at WileyPLUS
●When a planet or satellite with mass mmoves in a circular
orbit with radius r, its potential energy Uand kinetic energy K
are given by
UGMm
r and KGMm
2r.
The mechanical energy EKUis then
For an elliptical orbit of semimajor axis a,
EGMm
2a.
EGMm
2r.
Key Ideas
13-7 SATELLITES: ORBITS AND ENERGY
After reading this module, you should be able to . . .
13.22 For a satellite in a circular orbit around an astronomical
body, calculate the gravitational potential energy, the kinetic
energy, and the total energy.
13.23 For a satellite in an elliptical orbit, calculate the total
energy.
Learning Objectives
Satellites: Orbits and Energy
As a satellite orbits Earth in an elliptical path, both its speed, which fixes its kinetic
energy K, and its distance from the center of Earth, which fixes its gravitational po-
tential energy U, fluctuate with fixed periods. However, the mechanical energy Eof
the satellite remains constant. (Since the satellite’s mass is so much smaller than
Earth’s mass, we assign Uand Efor the Earth–satellite system to the satellite alone.)
372 CHAPTER 13 GRAVITATION
Figure 13-16 The variation of kinetic energy
K, potential energy U, and total energy E
with radius rfor a satellite in a circular or-
bit. For any value of r, the values of Uand
Eare negative, the value of Kis positive,
and EK.As r:, all three energy
curves approach a value of zero.
Energy
r
0
K(r)
E(r)
U(r)
This is a plot of a
satellite's energies
versus orbit radius.
The kinetic energy
is positive.
The potential energy
and total energy
are negative.
Checkpoint 5
In the figure here, a space shuttle is initially in a cir-
cular orbit of radius rabout Earth.At point P,
the pilot briefly fires a forward-pointing thruster
to decrease the shuttle’s kinetic energy Kand me-
chanical energy E. (a) Which of the dashed ellipti-
cal orbits shown in the figure will the shuttle then
take? (b) Is the orbital period Tof the shuttle (the
time to return to P) then greater than, less than, or
the same as in the circular orbit?
r
P
1
2
The potential energy of the system is given by Eq. 13-21:
(with U0 for infinite separation). Here ris the radius of the satellite’s orbit,
assumed for the time being to be circular, and Mand mare the masses of Earth
and the satellite, respectively.
To find the kinetic energy of a satellite in a circular orbit, we write Newton’s
second law (Fma) as
(13-37)
where v2/ris the centripetal acceleration of the satellite.Then, from Eq. 13-37, the
kinetic energy is
(13-38)
which shows us that for a satellite in a circular orbit,
(circular orbit). (13-39)
The total mechanical energy of the orbiting satellite is
or (circular orbit). (13-40)
This tells us that for a satellite in a circular orbit, the total energy Eis the negative of
the kinetic energy K:
EK(circular orbit). (13-41)
For a satellite in an elliptical orbit of semimajor axis a, we can substitute afor rin
Eq. 13-40 to find the mechanical energy:
(elliptical orbit). (13-42)
Equation 13-42 tells us that the total energy of an orbiting satellite de-
pends only on the semimajor axis of its orbit and not on its eccentricity e.For
example, four orbits with the same semimajor axis are shown in Fig. 13-15; the
same satellite would have the same total mechanical energy Ein all four or-
bits. Figure 13-16 shows the variation of K,U, and Ewith rfor a satellite mov-
ing in a circular orbit about a massive central body. Note that as ris increased,
the kinetic energy (and thus also the orbital speed) decreases.
EGMm
2a
EGMm
2r
EKUGMm
2rGMm
r
KU
2
K1
2mv2GMm
2r,
GMm
r2mv2
r,
UGMm
r
Figure 13-15 Four orbits with different eccen-
tricities eabout an object of mass M.All
four orbits have the same semimajor axis a
and thus correspond to the same total me-
chanical energy E.
e =
0
0.5
0.8
0.9
M
373
13-7 SATELLITES: ORBITS AND ENERGY
KEY IDEA
On the launchpad, the ball is not in orbit and thus Eq. 13-40
does not apply. Instead, we must find E0K0U0, where
K0is the ball’s kinetic energy and U0is the gravitational po-
tential energy of the ball–Earth system.
Calculations: To find U0, we use Eq. 13-21 to write
4.51 108J451 MJ.
The kinetic energy K0of the ball is due to the ball’s motion
with Earth’s rotation.You can show that K0is less than 1 MJ,
which is negligible relative to U0. Thus, the mechanical en-
ergy of the ball on the launchpad is
E0K0U00451 MJ 451 MJ. (Answer)
The increase in the mechanical energy of the ball from
launchpad to orbit is
EEE0(214 MJ) (451 MJ)
237 MJ. (Answer)
This is worth a few dollars at your utility company.
Obviously the high cost of placing objects into orbit is not
due to their required mechanical energy.
(6.67 1011 Nm2/kg2)(5.98 1024 kg)(7.20 kg)
6.37 106 m
U0GMm
R
Sample Problem 13.05 Mechanical energy of orbiting bowling ball
A playful astronaut releases a bowling ball, of mass m
7.20 kg, into circular orbit about Earth at an altitude hof
350 km.
(a) What is the mechanical energy Eof the ball in its orbit?
KEY IDEA
We can get Efrom the orbital energy, given by Eq. 13-40
(EGMm/2r), if we first find the orbital radius r. (It is
not simply the given altitude.)
Calculations: The orbital radius must be
rRh6370 km 350 km 6.72 106m,
in which Ris the radius of Earth. Then, from Eq. 13-40 with
Earth mass M5.98 1024 kg,the mechanical energy is
2.14 108J214 MJ. (Answer)
(b) What is the mechanical energy E0of the ball on the
launchpad at the Kennedy Space Center (before launch)?
From there to the orbit, what is the change Ein the ball’s
mechanical energy?
(6.67 1011 Nm2/kg2)(5.98 1024 kg)(7.20 kg)
(2)(6.72 106 m)
EGMm
2r
ence of the initial circular orbit to the initial period of the or-
bit. Thus, just after the thruster is fired, the kinetic energy is
1.0338 1011 J.
1
2(4.50 103 kg)(0.96)2
2p (8.00 106 m)
7.119 103 s
2
K1
2mv21
2m(0.96v0)21
2m(0.96)2
2pr
T
0
2
Sample Problem 13.06 Transforming a circular orbit into an elliptical orbit
A spaceship of mass m4.50 103kg is in a circular Earth
orbit of radius r8.00 106m and period T0118.6 min
7.119 103s when a thruster is fired in the forward direction
to decrease the speed to 96.0% of the original speed.What is
the period Tof the resulting elliptical orbit (Fig.13-17)?
KEY IDEAS
(1) The orbit of an elliptical orbit is related to the semima-
jor axis aby Kepler’s third law, written as Eq. 13-34 (
4p2r3/GM) but with areplacing r. (2) The semimajor axis a
is related to the total mechanical energy Eof the ship by
Eq. 13-42 (EGMm/2a), in which Earth’s mass is M
5.98 1024 kg. (3) The potential energy of the ship at radius
rfrom Earth’s center is given by Eq. 13-21 (UGMm/r).
Calculations: Looking over the Key Ideas, we see that we
need to calculate the total energy Eto find the semimajor
axis a, so that we can then determine the period of the ellipti-
cal orbit. Let’s start with the kinetic energy, calculating it just
after the thruster is fired.The speed vjust then is 96% of the
initial speed v0, which was equal to the ratio of the circumfer-
T2
Figure 13-17 At point Pa
thruster is fired, changing a
ship’s orbit from circular to
elliptical.
r
M
P
374 CHAPTER 13 GRAVITATION
Einstein and Gravitation
Principle of Equivalence
Albert Einstein once said: “I was . . . in the patent office at Bern when all of a
sudden a thought occurred to me: ‘If a person falls freely, he will not feel his
own weight.’ I was startled. This simple thought made a deep impression on me.
It impelled me toward a theory of gravitation.”
Thus Einstein tells us how he began to form his general theory of relativity.
The fundamental postulate of this theory about gravitation (the gravitating of
objects toward each other) is called the principle of equivalence, which says that
gravitation and acceleration are equivalent. If a physicist were locked up in a
small box as in Fig. 13-18, he would not be able to tell whether the box was at
(b)(a)
aa
Additional examples, video, and practice available at WileyPLUS
13-8 EINSTEIN AND GRAVITATION
After reading this module, you should be able to . . .
13.24 Explain Einstein’s principle of equivalence.
13.25 Identify Einstein’s model for gravitation as being due to
the curvature of spacetime.
●Einstein pointed out that gravitation and acceleration are equivalent. This principle of equivalence led him to a theory of gravi-
tation (the general theory of relativity) that explains gravitational effects in terms of a curvature of space.
Learning Objectives
Key Idea
Just after the thruster is fired, the ship is still at orbital
radius r, and thus its gravitational potential energy is
We can now find the semimajor axis by rearranging Eq. 13-42,
substituting afor r, and then substituting in our energy results:
7.418 106m.
(6.67 1011 Nm2/kg2)(5.98 1024 kg)(4.50 103kg)
2(1.0338 1011 J2.2436 1011 J)
aGMm
2E GMm
2(KU)
2.2436 1011 J.
(6.67 1011 Nm2/kg2)(5.98 1024 kg)(4.50 103kg)
8.00 106m
UGMm
r
OK, one more step to go. We substitute afor rin Eq. 13-34
and then solve for the period T, substituting our result for a:
(Answer)
This is the period of the elliptical orbit that the ship takes af-
ter the thruster is fired. It is less than the period T0for the cir-
cular orbit for two reasons. (1) The orbital path length is now
less. (2) The elliptical path takes the ship closer to Earth
everywhere except at the point of firing (Fig. 13-17). The re-
sulting decrease in gravitational potential energy increases
the kinetic energy and thus also the speed of the ship.
6.356 103s106 min.
4p2(7.418 106m)3
(6.67 1011 Nm2/kg2)(5.98 1024 kg)
1/2
T
4p2a3
GM
1/2
Figure 13-18 (a) A physicist in a box rest-
ing on Earth sees a cantaloupe falling
with acceleration a9.8 m/s2.(b) If he
and the box accelerate in deep space at
9.8 m/s2, the cantaloupe has the same
acceleration relative to him. It is not
possible, by doing experiments within
the box, for the physicist to tell which
situation he is in. For example, the plat-
form scale on which he stands reads
the same weight in both situations.
375
13-8 EINSTEIN AND GRAVITATION
Figure 13-19 (a) Two objects moving along lines of longitude toward the south pole converge
because of the curvature of Earth’s surface. (b) Two objects falling freely near Earth move
along lines that converge toward the center of Earth because of the curvature of space
near Earth. (c) Far from Earth (and other masses), space is flat and parallel paths remain
parallel. Close to Earth, the parallel paths begin to converge because space is curved by
Earth’s mass.
Earth
Converging
paths
Flat space
far from
Earth
Parallel paths
Curved space
near Earth
S
N
Equator
(a) (b)(c)
S
C
rest on Earth (and subject only to Earth’s gravitational force), as in Fig. 13-18a,
or accelerating through interstellar space at 9.8 m/s2(and subject only to the
force producing that acceleration), as in Fig. 13-18b. In both situations he would
feel the same and would read the same value for his weight on a scale. Moreover,
if he watched an object fall past him, the object would have the same acceleration
relative to him in both situations.
Curvature of Space
We have thus far explained gravitation as due to a force between masses. Einstein
showed that, instead, gravitation is due to a curvature of space that is caused by
the masses. (As is discussed later in this book, space and time are entangled, so
the curvature of which Einstein spoke is really a curvature of spacetime, the
combined four dimensions of our universe.)
Picturing how space (such as vacuum) can have curvature is difficult. An
analogy might help: Suppose that from orbit we watch a race in which two boats
begin on Earth’s equator with a separation of 20 km and head due south
(Fig. 13-19a). To the sailors, the boats travel along flat, parallel paths. However,
with time the boats draw together until, nearer the south pole, they touch. The
sailors in the boats can interpret this drawing together in terms of a force acting
on the boats. Looking on from space, however, we can see that the boats draw
together simply because of the curvature of Earth’s surface. We can see this
because we are viewing the race from “outside” that surface.
Figure 13-19bshows a similar race: Two horizontally separated apples are
dropped from the same height above Earth. Although the apples may appear to
travel along parallel paths, they actually move toward each other because they
both fall toward Earth’s center. We can interpret the motion of the apples in
terms of the gravitational force on the apples from Earth. We can also interpret
the motion in terms of a curvature of the space near Earth, a curvature due to the
presence of Earth’s mass. This time we cannot see the curvature because we
cannot get “outside” the curved space, as we got “outside” the curved Earth in
the boat example. However, we can depict the curvature with a drawing like Fig.
13-19c; there the apples would move along a surface that curves toward Earth
because of Earth’s mass.
When light passes near Earth, the path of the light bends slightly because of
the curvature of space there, an effect called gravitational lensing. When light
passes a more massive structure, like a galaxy or a black hole having large mass,
its path can be bent more. If such a massive structure is between us and a quasar
(an extremely bright, extremely distant source of light), the light from the quasar
376 CHAPTER 13 GRAVITATION
The Law of Gravitation Any particle in the universe attracts
any other particle with a gravitational force whose magnitude is
(Newton’s law of gravitation), (13-1)
where m1and m2are the masses of the particles, ris their separation,
and G(6.67 1011 Nm2/kg2) is the gravitational constant.
Gravitational Behavior of Uniform Spherical Shells
The gravitational force between extended bodies is found by
adding (integrating) the individual forces on individual particles
within the bodies. However, if either of the bodies is a uniform
spherical shell or a spherically symmetric solid, the net gravita-
tional force it exerts on an external object may be computed as if
all the mass of the shell or body were located at its center.
Superposition Gravitational forces obey the principle of su-
perposition; that is, if nparticles interact, the net force on a
particle labeled particle 1 is the sum of the forces on it from all the
other particles taken one at a time:
(13-5)
in which the sum is a vector sum of the forces on particle
1 from particles 2, 3,..., n. The gravitational force on aF
:
1
F
:
1i
F
:
1,net
n
i2
F
:
1i,
F
:
1,net
FGm1m2
r2
Review & Summary
particle from an extended body is found by dividing the body into
units of differential mass dm, each of which produces a differential
force on the particle, and then integrating to find the sum of
those forces:
(13-6)
Gravitational Acceleration The gravitational acceleration ag
of a particle (of mass m) is due solely to the gravitational force acting
on it.When the particle is at distance rfrom the center of a uniform,
spherical body of mass M, the magnitude Fof the gravitational force
on the particle is given by Eq.13-1.Thus, by Newton’s second law,
Fmag, (13-10)
which gives
. (13-11)
Free-Fall Acceleration and Weight Because Earth’s mass
is not distributed uniformly, because the planet is not perfectly
spherical, and because it rotates, the actual free-fall acceleration
of a particle near Earth differs slightly from the gravitational accel-
eration and the particle’s weight (equal to mg) differs from the
magnitude of the gravitational force on it as calculated by
Newton’s law of gravitation (Eq.13-1).
a
:
g,
g
:
agGM
r2
F
:
1dF
:
.
dF
:
can bend around the massive structure and toward us (Fig.13-20a).Then, because
the light seems to be coming to us from a number of slightly different directions
in the sky, we see the same quasar in all those different directions. In some situa-
tions, the quasars we see blend together to form a giant luminous arc, which is
called an Einstein ring (Fig. 13-20b).
Should we attribute gravitation to the curvature of spacetime due to the
presence of masses or to a force between masses? Or should we attribute it to
the actions of a type of fundamental particle called a graviton, as conjectured in
some modern physics theories? Although our theories about gravitation have
been enormously successful in describing everything from falling apples to plane-
tary and stellar motions, we still do not fully understand it on either the cosmo-
logical scale or the quantum physics scale.
Paths of light
from quasar
Apparent
quasar directions
Galaxy or
large black hole
Final paths
Earth detector
(a)(b)
Courtesy National Radio Astronomy Observatory
Figure 13-20 (a) Light from a distant quasar
follows curved paths around a galaxy or
a large black hole because the mass of the
galaxy or black hole has curved the adja-
cent space. If the light is detected,it ap-
pears to have originated along the back-
ward extensions of the final paths (dashed
lines). (b) The Einstein ring known as
MG11310456 on the computer screen of a
telescope.The source of the light (actually,
radio waves, which are a form of invisible
light) is far behind the large, unseen galaxy
that produces the ring; a portion of the
source appears as the two bright spots seen
along the ring.
377
QUESTIONS
Gravitation Within a Spherical Shell A uniform shell of
matter exerts no net gravitational force on a particle located inside
it. This means that if a particle is located inside a uniform solid
sphere at distance rfrom its center, the gravitational force exerted
on the particle is due only to the mass that lies inside a sphere of
radius r(the inside sphere).The force magnitude is given by
(13-19)
where Mis the sphere’s mass and Ris its radius.
Gravitational Potential Energy The gravitational potential
energy U(r) of a system of two particles, with masses Mand mand
separated by a distance r, is the negative of the work that would be
done by the gravitational force of either particle acting on the other
if the separation between the particles were changed from infinite
(very large) to r.This energy is
(gravitational potential energy). (13-21)
Potential Energy of a System If a system contains more
than two particles, its total gravitational potential energy Uis the
sum of the terms representing the potential energies of all the pairs.
As an example, for three particles, of masses m1,m2,and m3,
(13-22)
Escape Speed An object will escape the gravitational pull of
an astronomical body of mass Mand radius R(that is, it will reach
an infinite distance) if the object’s speed near the body’s surface is
at least equal to the escape speed, given by
(13-28)vA2GM
R.
U
Gm1m2
r12
Gm1m3
r13
Gm2m3
r23
.
UGMm
r
FGmM
R3r,
Kepler’s Laws The motion of satellites, both natural and artifi-
cial, is governed by these laws:
1. The law of orbits. All planets move in elliptical orbits with the
Sun at one focus.
2. The law of areas. A line joining any planet to the Sun sweeps
out equal areas in equal time intervals. (This statement is equiv-
alent to conservation of angular momentum.)
3. The law of periods. The square of the period Tof any planet is
proportional to the cube of the semimajor axis aof its orbit. For
circular orbits with radius r,
(law of periods), (13-34)
where Mis the mass of the attracting body—the Sun in the case
of the solar system. For elliptical planetary orbits, the semi-
major axis ais substituted for r.
Energy in Planetary Motion When a planet or satellite with
mass mmoves in a circular orbit with radius r, its potential energy
Uand kinetic energy Kare given by
(13-21, 13-38)
The mechanical energy EKUis then
(13-40)
For an elliptical orbit of semimajor axis a,
(13-42)
Einstein’s View of Gravitation Einstein pointed out that gravi-
tation and acceleration are equivalent. This principle of equivalence
led him to a theory of gravitation (the general theory of relativity) that
explains gravitational effects in terms of a curvature of space.
EGMm
2a.
EGMm
2r.
UGMm
r and KGMm
2r.
T2
4p2
GM
r3
Questions
1In Fig. 13-21, a central particle of
mass Mis surrounded by a square ar-
ray of other particles, separated by ei-
ther distance dor distance d/2 along
the perimeter of the square.What are
the magnitude and direction of the
net gravitational force on the central
particle due to the other particles?
2Figure 13-22 shows three
arrangements of the same identical
particles, with three of them placed
on a circle of radius 0.20 m and the
fourth one placed at the center of
the circle. (a) Rank the arrange-
ments according to the magnitude of
the net gravitational force on the
central particle due to the other
three particles, greatest first. (b)
Rank them according to the gravitational potential energy of the
four-particle system, least negative first.
3In Fig. 13-23, a central particle is surrounded by two circular
M
M
M
2M
7M
3M
5M
4M
4M
5M
7M
2M
Figure 13-21 Question 1.
(
a
) (
b
) (
c
)
Figure 13-22 Question 2.
rings of particles, at radii rand R,
with Rr. All the particles have
mass m. What are the magnitude
and direction of the net gravita-
tional force on the central particle
due to the particles in the rings?
4In Fig. 13-24, two particles, of
masses mand 2m, are fixed in place
on an axis. (a) Where on the axis can
a third particle of mass 3mbe placed
(other than at infinity) so that the
net gravitational force on it from the
first two particles is zero: to the left
of the first two particles, to their
right, between them but closer to
the more massive particle, or be-
tween them but closer to the less
massive particle? (b) Does the an-
swer change if the third particle has, instead, a mass of 16m? (c) Is
there a point off the axis (other than infinity) at which the net force
on the third particle would be zero?
Figure 13-23 Question 3.
Figure 13-24 Question 4.
m2m
378 CHAPTER 13 GRAVITATION
a
b
16 h
c
d
24 h
e
f
48 h
Figure 13-30 Question 11.
a2m m b c
d d d d
Figure 13-31 Question 12.
Module 13-1 Newton’s Law of Gravitation
•1 A mass Mis split into two parts, mand Mm, which are
then separated by a certain distance. What ratio m/Mmaximizes
the magnitude of the gravitational force between the parts?
ILW
5Figure 13-25 shows three situations involving a point particle P
with mass mand a spherical shell with a uniformly distributed
mass M. The radii of the shells are given. Rank the situations ac-
cording to the magnitude of the gravitational force on particle P
due to the shell, greatest first.
Figure 13-25 Question 5.
R
2RR/2
PP
P
(
a
)(
b
)(
c
)
d
R1R2R4
r
ag
R3
1
2
3
4
Figure 13-27 Question 8.
9Figure 13-28 shows three parti-
cles initially fixed in place, with B
and Cidentical and positioned sym-
metrically about the yaxis, at dis-
tance dfrom A. (a) In what direction
is the net gravitational force F
:
net on
A? (b) If we move Cdirectly away
from the origin, does F
:
net change in
direction? If so, how and what is the
limit of the change?
10 Figure 13-29 shows six paths by
which a rocket orbiting a moon might
move from point ato point b.Rank the
paths according to (a) the correspon-
ding change in the gravitational poten-
tial energy of the rocket–moon system
and (b) the net work done on the
rocket by the gravitational force from
the moon,greatest first.
11 Figure 13-30 shows three uniform spherical planets that are
identical in size and mass. The periods of rotation Tfor the planets
are given, and six lettered points are indicated—three points are
on the equators of the planets and three points are on the north
poles. Rank the points according to the value of the free-fall accel-
eration gat them, greatest first.
Figure 13-28 Question 9.
y
CB
d d
Ax
θ θ
3
2
15
6
a
b
4
Figure 13-29 Question 10.
6In Fig. 13-26, three particles are
fixed in place. The mass of Bis
greater than the mass of C. Can a
fourth particle (particle D) be placed
somewhere so that the net gravita-
tional force on particle Afrom parti-
cles B, C, and Dis zero? If so, in
which quadrant should it be placed
and which axis should it be near?
7Rank the four systems of equal-
mass particles shown in Checkpoint 2
according to the absolute value of the gravitational potential energy
of the system,greatest first.
8Figure 13-27 gives the gravitational acceleration agfor four planets
as a function of the radial distance rfrom the center of the planet, start-
ing at the surface of the planet (at radius R1,R2,R3, or R4). Plots 1 and 2
coincide for rR2;plots 3 and 4 coincide for rR4.Rank the four plan-
ets according to (a) mass and (b) mass per unit volume,greatest first.
y
x
d
d
A C
B
Figure 13-26 Question 6.
12 In Fig. 13-31, a particle of mass m(which is not shown) is to
be moved from an infinite distance to one of the three possible
locations a, b, and c.Two other particles, of masses mand 2m,are
already fixed in place on the axis, as shown. Rank the three pos-
sible locations according to the work done by the net gravita-
tional force on the moving particle due to the fixed particles,
greatest first.
•2 Moon effect. Some people believe that the Moon con-
trols their activities. If the Moon moves from being directly on the
opposite side of Earth from you to being directly overhead, by
what percent does (a) the Moon’s gravitational pull on you
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
20.0 cm is formed by four spheres of masses
m15.00 g, m23.00 g, m31.00 g, and
m45.00 g. In unit-vector notation, what is
the net gravitational force from them on a
central sphere with mass m52.50 g?
•7 One dimension. In Fig. 13-33, two
point particles are fixed on an xaxis sepa-
rated by distance d. Particle Ahas mass mA
and particle Bhas mass 3.00mA. A third
particle C, of mass 75.0mA, is to be placed
on the xaxis and near particles Aand B.In
terms of distance d, at what xcoordinate
should Cbe placed so that the net gravita-
tional force on particle Afrom particles B
and Cis zero?
•8 In Fig. 13-34, three 5.00 kg spheres are lo-
cated at distances d10.300 m and d20.400
m.What are the (a) magnitude and (b) direction (relative to the positive
direction of the xaxis) of the net gravitational force on sphere Bdue to
spheres Aand C?
increase and (b) your weight (as measured on a scale) decrease?
Assume that the Earth–Moon (center-to-center) distance is
3.82 108m and Earth’s radius is 6.37 106m.
•3 What must the separation be between a 5.2 kg particle
and a 2.4 kg particle for their gravitational attraction to have
a magnitude of 2.3 1012 N?
•4 The Sun and Earth each exert a gravitational force on the
Moon. What is the ratio FSun/FEarth of these two forces? (The aver-
age Sun–Moon distance is equal to the Sun–Earth distance.)
•5 Miniature black holes. Left over from the big-bang beginning
of the universe, tiny black holes might still wander through the
universe. If one with a mass of 1 1011 kg (and a radius of only
11016 m) reached Earth, at what distance from your head
would its gravitational pull on you match that of Earth’s?
Module 13-2 Gravitation and the Principle of Superposition
•6 In Fig. 13-32, a square of edge length
SSM
fixed in place at x0.20 m on the xaxis and particle B, with a
mass of 1.0 kg, is fixed in place at the origin. Particle C(not shown)
can be moved along the xaxis, between particle Band x.
Figure 13-37bshows the xcomponent Fnet,xof the net gravitational
force on particle Bdue to particles Aand C, as a function of posi-
tion xof particle C. The plot actually extends to the right, ap-
proaching an asymptote of 4.17 1010 N as x:.What are the
masses of (a) particle Aand (b) particle C?
379
PROBLEMS
tance d, at what (a) xcoordinate and (b) ycoordinate should parti-
cle Dbe placed so that the net gravitational force on particle A
from particles B,C,and Dis zero?
••11 As seen in Fig. 13-36, two
spheres of mass mand a third sphere
of mass Mform an equilateral trian-
gle, and a fourth sphere of mass m4is
at the center of the triangle. The net
gravitational force on that central
sphere from the three other spheres is
zero. (a) What is Min terms of m? (b)
If we double the value of m4, what
then is the magnitude of the net gravi-
tational force on the central sphere?
••12 In Fig. 13-37a, particle Ais
m1
m3m
4
m5
m2
y
x
Figure 13-32
Problem 6.
y
x
d
A B
Figure 13-33
Problem 7.
Figure 13-34 Problem 8.
y
x
d2
d1
A
B C
m m
m4
M
Figure 13-36
Problem 11.
y
x
(
a
)(
b
)
A B 0
0
x (m)
Fnet,x
0.2 0.4 0.6 0.8
Figure 13-37 Problem 12.
••13 Figure 13-38 shows a spherical
hollow inside a lead sphere of radius
R4.00 cm; the surface of the hol-
low passes through the center of the
sphere and “touches” the right
side of the sphere. The mass of the
sphere before hollowing was M
2.95 kg. With what gravitational
force does the hollowed-out lead sphere attract a small sphere
of mass m0.431 kg that lies at a distance d9.00 cm from
the center of the lead sphere, on the straight line connecting the
centers of the spheres and of the hollow?
••14 Three point particles are
fixed in position in an xy plane. Two of
them, particle Aof mass 6.00 g and par-
ticle Bof mass 12.0 g, are shown in Fig.
13-39, with a separation of dAB 0.500
m at angle u30°. Particle C, with mass
8.00 g, is not shown. The net gravita-
tional force acting on particle Adue to
particles Band Cis 2.77 1014 N at
an angle of 163.8° from the positive direction of the xaxis. What
are (a) the xcoordinate and (b) the ycoordinate of particle C?
•••15 Three dimensions. Three point particles are fixed in place
in an xyz coordinate system. Particle A, at the origin, has mass mA.
m
R
d
Figure 13-38 Problem 13.
y
x
A
B
dAB
θ
Figure 13-39 Problem 14.
•9 We want to position a space probe along a line that
extends directly toward the Sun in order to monitor solar flares. How
far from Earth’s center is the point on the
line where the Sun’s gravitational pull on
the probe balances Earth’s pull?
••10 Two dimensions. In Fig. 13-35,
three point particles are fixed in place in
an xy plane. Particle Ahas mass mA, par-
ticle Bhas mass 2.00mA, and particle C
has mass 3.00mA. A fourth particle D,
with mass 4.00mA, is to be placed near
the other three particles. In terms of dis-
WWWSSM
y
x
B
AC
d
1.5d
Figure 13-35 Problem 10.
380 CHAPTER 13 GRAVITATION
6345 km
25 km
3490 km
Core, 1.93 1024 kg
Mantle, 4.01 1024 kg
Crust, 3.94 1022 kg
Figure 13-42 Problem 27.
Particle B, at xyz coordinates (2.00d, 1.00d, 2.00d), has mass 2.00mA,
and particle C, at coordinates (1.00d, 2.00d,3.00d), has mass
3.00mA. A fourth particle D, with mass 4.00mA, is to be placed near
the other particles. In terms of distance d, at what (a) x, (b) y, and (c) z
coordinate should Dbe placed so that the net gravitational force on
Afrom B,C, and Dis zero?
•••16 In Fig. 13-40, a particle
shells, when the particle is located at radial distance (a) a, (b) b,
and (c) c.
••25 A solid sphere has a uniformly distributed mass of 1.0 104
kg and a radius of 1.0 m.What is the magnitude of the gravitational
force due to the sphere on a particle of mass mwhen the particle is
located at a distance of (a) 1.5 m and (b) 0.50 m from the center of
the sphere? (c) Write a general expression for the magnitude
of the gravitational force on the particle at a distance r1.0 m
from the center of the sphere.
••26 A uniform solid sphere of radius Rproduces a gravitational
acceleration of agon its surface.At what distance from the sphere’s
center are there points (a) inside and (b) outside the sphere where
the gravitational acceleration is ag/3?
••27 Figure 13-42 shows, not to scale, a cross section through
the interior of Earth. Rather than being uniform throughout,
Earth is divided into three zones: an outer crust, amantle, and an
inner core. The dimensions of these zones and the masses con-
tained within them are shown on the figure. Earth has a total
mass of 5.98 1024 kg and a radius of 6370 km. Ignore rotation
and assume that Earth is spherical. (a) Calculate agat the sur-
face. (b) Suppose that a bore hole (the Mohole) is driven to the
crust–mantle interface at a depth of 25.0 km; what would be the
value of agat the bottom of the hole? (c) Suppose that Earth
were a uniform sphere with the same total mass and size. What
would be the value of agat a depth of 25.0 km? (Precise mea-
surements of agare sensitive probes of the interior structure of
Earth, although results can be clouded by local variations in
mass distribution.)
of mass m1 0.67 kg is a dis-
tance d23 cm from one end of
a uniform rod with length L
3.0 m and mass M 5.0 kg.What
is the magnitude of the gravita-
tional force F
:on the particle from the rod?
Module 13-3 Gravitation Near Earth’s Surface
•17 (a) What will an object weigh on the Moon’s surface if it
weighs 100 N on Earth’s surface? (b) How many Earth radii must
this same object be from the center of Earth if it is to weigh the
same as it does on the Moon?
•18 Mountain pull. A large mountain can slightly affect the
direction of “down” as determined by a plumb line. Assume that
we can model a mountain as a sphere of radius R2.00 km and
density (mass per unit volume) 2.6 103kg/m3. Assume also that
we hang a 0.50 m plumb line at a distance of 3Rfrom the sphere’s
center and such that the sphere pulls horizontally on the lower
end. How far would the lower end move toward the sphere?
•19 At what altitude above Earth’s surface would the
gravitational acceleration be 4.9 m/s2?
•20 Mile-high building. In 1956, Frank Lloyd Wright proposed
the construction of a mile-high building in Chicago. Suppose the
building had been constructed. Ignoring Earth’s rotation, find
the change in your weight if you were to ride an elevator from the
street level, where you weigh 600 N, to the top of the building.
••21 Certain neutron stars (extremely dense stars) are
believed to be rotating at about 1 rev/s. If such a star has a radius of
20 km, what must be its minimum mass so that material on its sur-
face remains in place during the rapid rotation?
••22 The radius Rhand mass Mhof a black hole are related by
Rh2GMh/c2, where cis the speed of light.Assume that the gravi-
tational acceleration agof an object at a distance ro1.001Rhfrom
the center of a black hole is given by Eq. 13-11 (it is, for large black
holes).(a) In terms of Mh,find agat ro.(b) Does agat roincrease or de-
crease as Mhincreases? (c) What is agat rofor a very large black hole
whose mass is 1.55 1012 times the solar mass of 1.99 1030 kg?
(d) If an astronaut of height 1.70 m is at rowith her feet down, what is
the difference in gravitational acceleration between her head and
feet? (e) Is the tendency to stretch the astronaut severe?
••23 One model for a certain planet has a core of radius Rand mass
Msurrounded by an outer shell of inner radius R, outer radius 2R,
and mass 4M. If M4.1 1024 kg and
R6.0 106m, what is the gravitational
acceleration of a particle at points (a) R
and (b) 3Rfrom the center of the planet?
Module 13-4 Gravitation Inside Earth
•24 Two concentric spherical shells with
uniformly distributed masses M1and M2
are situated as shown in Fig. 13-41. Find
the magnitude of the net gravitational
force on a particle of mass m, due to the
ILW
SSM
Figure 13-40 Problem 16.
dL
dr
rdm
m1
••28 Assume a planet is a uniform sphere of radius Rthat
(somehow) has a narrow radial tunnel through its center
(Fig. 13-7). Also assume we can position an apple anywhere
along the tunnel or outside the sphere. Let FRbe the magnitude
of the gravitational force on the apple when it is located at
the planet’s surface. How far
from the surface is there a point
where the magnitude is FRif
we move the apple (a) away
from the planet and (b) into
the tunnel?
Module 13-5 Gravitational
Potential Energy
•29 Figure 13-43 gives the po-
tential energy function U(r) of a
projectile, plotted outward from
1
2
M2
M1c
b
a
Figure 13-41 Problem 24. Figure 13-43 Problems 29 and 34.
0
–1
–2
–3
–4
–5
r
Rs
U (109J)
of 5.0 1023 kg, a radius of 3.0 106m, and no
atmosphere. A 10 kg space probe is to be
launched vertically from its surface. (a) If the probe is launched
with an initial energy of 5.0 107J, what will be its kinetic energy
when it is 4.0 106m from the center of Zero? (b) If the probe is
to achieve a maximum distance of 8.0 106m from the center of
Zero, with what initial kinetic energy must it be launched from the
surface of Zero?
••37 The three spheres in Fig. 13-45, with masses mA80 g,
381
PROBLEMS
Module 13-6 Planets and Satellites: Kepler’s Laws
•43 (a) What linear speed must an Earth satellite have to be in
a circular orbit at an altitude of 160 km above Earth’s surface?
(b) What is the period of revolution?
•44 A satellite is put in a circular orbit about Earth with a radius
equal to one-half the radius of the Moon’s orbit.What is its period
of revolution in lunar months? (A lunar month is the period of rev-
olution of the Moon.)
•45 The Martian satellite Phobos travels in an approximately cir-
cular orbit of radius 9.4 106m with a period of 7 h 39 min.
Calculate the mass of Mars from this information.
•46 The first known collision between space debris and a func-
tioning satellite occurred in 1996: At an altitude of 700 km, a year-
old French spy satellite was hit by a piece of an Ariane rocket. A
stabilizing boom on the satellite was demolished, and the satellite
the surface of a planet of radius Rs.What least kinetic energy is re-
quired of a projectile launched at the surface if the projectile is to
“escape” the planet?
•30 In Problem 1, what ratio m/Mgives the least gravitational
potential energy for the system?
•31 The mean diameters of Mars and Earth are 6.9 103km
SSM
••38 In deep space, sphere Aof mass 20 kg is located at the origin
of an xaxis and sphere Bof mass 10 kg is located on the axis at x
0.80 m. Sphere Bis released from rest while sphere Ais held at the
origin. (a) What is the gravitational potential energy of the two-
sphere system just as Bis released? (b) What is the kinetic energy
of Bwhen it has moved 0.20 m toward A?
1.0 1010 m. They each have a mass of 1.0 1030 kg and a radius
of 1.0 105m.They are initially at rest with respect to each other.
As measured from that rest frame, how fast are they moving when
(a) their separation has decreased to one-half its initial value and
(b) they are about to collide?
••42 Figure 13-46ashows a particle Athat can be moved
along a yaxis from an infinite distance to the origin.That origin lies
at the midpoint between particles Band C, which have identical
masses, and the yaxis is a perpendicular bisector between them.
Distance Dis 0.3057 m. Figure 13-46bshows the potential energy
Uof the three-particle system as a function of the position of parti-
cle Aalong the yaxis. The curve actually extends rightward and ap-
proaches an asymptote of 2.7 1011 J as y:. What are the
masses of (a) particles Band Cand (b) particle A?
and 1.3 104km, respectively. The mass of Mars is 0.11 times
Earth’s mass. (a) What is the ratio of the mean density (mass per
unit volume) of Mars to that of Earth? (b) What is the value of the
gravitational acceleration on Mars? (c) What is the escape speed
on Mars?
•32 (a) What is the gravitational potential energy of the two-
particle system in Problem 3? If you triple the separation between
the particles, how much work is done (b) by the gravitational force
between the particles and (c) by you?
•33 What multiple of the energy needed to escape from Earth gives
the energy needed to escape from (a) the Moon and (b) Jupiter?
•34 Figure 13-43 gives the potential energy function U(r) of
a projectile, plotted outward from the surface of a planet of
radius Rs. If the projectile is launched radially outward from the
surface with a mechanical energy of 2.0 109J, what are (a) its
kinetic energy at radius r1.25Rsand (b) its turning point (see
Module 8-3) in terms of Rs?
••35 Figure 13-44 shows four particles, each
of mass 20.0 g, that form a square with an edge
length of d0.600 m. If dis reduced to 0.200 m,
what is the change in the gravitational potential
energy of the four-particle system?
••36 Zero, a hypothetical planet, has a mass
d
d
Figure 13-44
Problem 35.
L
d d
A
BC
Figure 13-45 Problem 37.
mB10 g, and mC20 g, have their centers on a common line,
with L12 cm and d4.0 cm.You move sphere Balong the line
until its center-to-center separation from Cis d4.0 cm. How
much work is done on sphere B(a) by you and (b) by the net gravi-
tational force on Bdue to spheres Aand C?
Figure 13-46 Problem 42.
A
CB
D D
y
x
(
a
)(
b
)
0
0.5 1 1.5 2
–1
–2
–3
y (m)
U (10–10 J)
••39 (a) What is the escape speed on a spherical asteroid
whose radius is 500 km and whose gravitational acceleration at the
surface is 3.0 m/s2? (b) How far from the surface will a particle go if
it leaves the asteroid’s surface with a radial speed of 1000 m/s? (c)
With what speed will an object hit the asteroid if it is dropped from
1000 km above the surface?
••40 A projectile is shot directly away from Earth’s surface.
Neglect the rotation of Earth. What multiple of Earth’s radius RE
gives the radial distance a projectile reaches if (a) its initial speed is
0.500 of the escape speed from Earth and (b) its initial kinetic en-
ergy is 0.500 of the kinetic energy required to escape Earth? (c)
What is the least initial mechanical energy required at launch if the
projectile is to escape Earth?
••41 Two neutron stars are separated by a distance of
SSM
SSM
382 CHAPTER 13 GRAVITATION
Name a(108m) T(days)
Io 4.22 1.77
Europa 6.71 3.55
Ganymede 10.7 7.16
Callisto 18.8 16.7
Figure 13-48 Problem 56. A tiny moon (at right) orbits
asteroid 243 Ida.
Courtesy NASA
70
0
–70
Line of sight velocity (m/s)
1500 days
Time
Figure 13-49 Problem 58.
(a) Plot log a(yaxis) against log T(xaxis) and show that you get a
straight line. (b) Measure the slope of the line and compare it with
the value that you expect from Kepler’s third law. (c) Find the mass
of Jupiter from the intercept of this line with the yaxis.
••56 In 1993 the spacecraft Galileo sent an image (Fig. 13-48) of
asteroid 243 Ida and a tiny orbiting moon (now known as Dactyl), the
first confirmed example of an asteroid–moon system. In the image,
the moon, which is 1.5 km wide, is 100 km from the center of the as-
teroid,which is 55 km long.Assume the moon’s orbit is circular with a
period of 27 h. (a) What is the mass of the asteroid? (b) The volume
of the asteroid, measured from the Galileo images, is 14 100 km3.
What is the density (mass per unit volume) of the asteroid?
was sent spinning out of control. Just before the collision and in
kilometers per hour, what was the speed of the rocket piece rela-
tive to the satellite if both were in circular orbits and the collision
was (a) head-on and (b) along perpendicular paths?
•47 The Sun, which is 2.2 1020 m from the center
of the Milky Way galaxy, revolves around that center once every
2.5 108years. Assuming each star in the Galaxy has a mass equal
to the Sun’s mass of 2.0 1030 kg, the stars are distributed uni-
formly in a sphere about the galactic center, and the Sun is at the
edge of that sphere, estimate the number of stars in the Galaxy.
•48 The mean distance of Mars from the Sun is 1.52 times that of
Earth from the Sun. From Kepler’s law of periods, calculate the
number of years required for Mars to make one revolution around
the Sun; compare your answer with the value given in Appendix C.
•49 A comet that was seen in April 574 by Chinese astronomers
on a day known by them as the Woo Woo day was spotted again in
May 1994. Assume the time between observations is the period of
the Woo Woo day comet and its eccentricity is 0.9932.What are (a)
the semimajor axis of the comet’s orbit and (b) its greatest distance
from the Sun in terms of the mean orbital radius RPof Pluto?
•50 An orbiting satellite stays over a certain spot on the
equator of (rotating) Earth.What is the altitude of the orbit (called
ageosynchronous orbit)?
•51 A satellite, moving in an elliptical orbit, is 360 km above
Earth’s surface at its farthest point and 180 km above at its closest point.
Calculate (a) the semimajor axis and (b) the eccentricity of the orbit.
•52 The Sun’s center is at one focus of Earth’s orbit. How far
from this focus is the other focus, (a) in meters and (b) in terms of
the solar radius, 6.96 108m? The eccentricity is 0.0167, and the
semimajor axis is 1.50 1011 m.
••53 A 20 kg satellite has a circular orbit with a period of 2.4 h
and a radius of 8.0 106m around a planet of unknown mass. If
the magnitude of the gravitational acceleration on the surface of
the planet is 8.0 m/s2,what is the radius of the planet?
••54 Hunting a black hole.
Observations of the light from a
certain star indicate that it is part of
a binary (two-star) system.This visi-
ble star has orbital speed v270
km/s, orbital period T1.70 days,
and approximate mass m16Ms,
where Msis the Sun’s mass, 1.99
1030 kg.Assume that the visible star
and its companion star, which is
dark and unseen, are both in circu-
lar orbits (Fig. 13-47). What integer
multiple of Msgives the approxi-
mate mass m2of the dark star?
••55 In 1610, Galileo used his telescope to discover four moons
around Jupiter, with these mean orbital radii aand periods T:
SSM
WWWSSM
Figure 13-47 Problem 54.
O
m1m2
r2
r1
••57 In a certain binary-star system, each star has the same
mass as our Sun, and they revolve about their center of mass. The
distance between them is the same as the distance between Earth
and the Sun.What is their period of revolution in years?
•••58 The presence of an unseen planet orbiting a distant star
can sometimes be inferred from the motion of the star as we see
it. As the star and planet orbit the center of mass of the
star–planet system, the star moves toward and away from us with
what is called the line of sight velocity, a motion that can be de-
tected. Figure 13-49 shows a graph of the line of sight velocity ver-
sus time for the star 14 Herculis. The star’s mass is believed to be
0.90 of the mass of our Sun.Assume that only one planet orbits the
star and that our view is along the plane of the orbit.Then approxi-
mate (a) the planet’s mass in terms of Jupiter’s mass mJand (b) the
planet’s orbital radius in terms of Earth’s orbital radius rE.
ILW
•••59 Three identical stars of mass Mform an equilateral triangle
that rotates around the triangle’s center as the stars move in a com-
mon circle about that center. The triangle has edge length L. What
is the speed of the stars?
383
PROBLEMS
Module 13-7 Satellites: Orbits and Energy
•60 In Fig. 13-50, two satellites, Aand B,
both of mass m125 kg, move in the
same circular orbit of radius r7.87 106
m around Earth but in opposite senses of
rotation and therefore on a collision
course. (a) Find the total mechanical en-
ergy EAEBof the two satellites
Earth system before the collision. (b) If
the collision is completely inelastic so that
the wreckage remains as one piece of tan-
gled material (mass 2m), find the total
mechanical energy immediately after the collision. (c) Just after the
collision, is the wreckage falling directly toward Earth’s center or or-
biting around Earth?
•61 (a) At what height above Earth’s surface is the energy re-
quired to lift a satellite to that height equal to the kinetic energy
required for the satellite to be in orbit at that height? (b) For
greater heights, which is greater,the energy for lifting or the kinetic
energy for orbiting?
•62 Two Earth satellites, Aand B, each of mass m, are to be
launched into circular orbits about Earth’s center. Satellite Ais
to orbit at an altitude of 6370 km. Satellite Bis to orbit at an
altitude of 19 110 km.The radius of Earth REis 6370 km. (a) What
is the ratio of the potential energy of satellite Bto that of satel-
lite A, in orbit? (b) What is the ratio of the kinetic energy of
satellite Bto that of satellite A, in orbit? (c) Which satellite has
the greater total energy if each has a mass of 14.6 kg? (d) By
how much?
•63 An asteroid, whose mass is 2.0 104times the
mass of Earth, revolves in a circular orbit around the Sun at a dis-
tance that is twice Earth’s distance from the Sun. (a) Calculate the
period of revolution of the asteroid in years. (b) What is the ratio of
the kinetic energy of the asteroid to the kinetic energy of Earth?
•64 A satellite orbits a planet of unknown mass in a circle of ra-
dius 2.0 107m. The magnitude of the gravitational force on the
satellite from the planet is F80 N. (a) What is the kinetic energy
of the satellite in this orbit? (b) What would Fbe if the orbit radius
were increased to 3.0 107m?
••65 A satellite is in a circular Earth orbit of radius r.The area A
enclosed by the orbit depends on r2because Apr2. Determine
how the following properties of the satellite depend on r: (a) pe-
riod, (b) kinetic energy, (c) angular momentum,and (d) speed.
••66 One way to attack a satellite in Earth orbit is to launch a
swarm of pellets in the same orbit as the satellite but in the oppo-
site direction. Suppose a satellite in a circular orbit 500 km above
Earth’s surface collides with a pellet having mass 4.0 g. (a) What is
the kinetic energy of the pellet in the reference frame of the satel-
lite just before the collision? (b) What is the ratio of this kinetic en-
ergy to the kinetic energy of a 4.0 g bullet from a modern army ri-
fle with a muzzle speed of 950 m/s?
•••67 What are (a) the speed and (b) the period of a 220 kg satel-
lite in an approximately circular orbit 640 km above the surface of
Earth? Suppose the satellite loses mechanical energy at the aver-
age rate of 1.4 105J per orbital revolution.Adopting the reason-
able approximation that the satellite’s orbit becomes a “circle of
slowly diminishing radius,” determine the satellite’s (c) altitude,(d)
speed, and (e) period at the end of its 1500th revolution. (f) What
WWWSSM
is the magnitude of the average retarding force on the satellite? Is
angular momentum around Earth’s center conserved for (g) the
satellite and (h) the satellite–Earth system (assuming that system
is isolated)?
•••68 Two small spaceships, each with mass m2000 kg,are in
the circular Earth orbit of Fig. 13-51, at an altitude hof 400 km.
Igor, the commander of one of the ships, arrives at any fixed point
in the orbit 90 s ahead of Picard, the
commander of the other ship. What
are the (a) period T0and (b) speed
v0of the ships? At point Pin
Fig. 13-51, Picard fires an instanta-
neous burst in the forward direction,
reducing his ship’s speed by 1.00%.
After this burst, he follows the ellip-
tical orbit shown dashed in the fig-
ure. What are the (c) kinetic
energy and (d) potential energy of
his ship immediately after the burst?
In Picard’s new elliptical orbit, what are (e) the total
energy E, (f) the semimajor axis a, and (g) the orbital period T?
(h) How much earlier than Igor will Picard return to P?
Module 13-8 Einstein and Gravitation
•69 In Fig. 13-18b, the scale on which the 60 kg physicist stands
reads 220 N. How long will the cantaloupe take to reach the floor if
the physicist drops it (from rest relative to himself) at a height of
2.1 m above the floor?
Additional Problems
70 The radius Rhof a black hole is the radius of a mathemati-
cal sphere, called the event horizon, that is centered on the black
hole. Information from events inside the event horizon cannot
reach the outside world. According to Einstein’s general theory of
relativity, Rh2GM/c2, where Mis the mass of the black hole and
cis the speed of light.
Suppose that you wish to study a black hole near it, at a radial
distance of 50Rh. However, you do not want the difference in gravi-
tational acceleration between your feet and your head to exceed
10 m/s2when you are feet down (or head down) toward the black
hole. (a) As a multiple of our Sun’s mass MS, approximately what is
the limit to the mass of the black hole you can tolerate at the given
radial distance? (You need to estimate your height.) (b) Is the limit
an upper limit (you can tolerate smaller masses) or a lower limit
(you can tolerate larger masses)?
71 Several planets (Jupiter, Saturn,
Uranus) are encircled by rings, perhaps
composed of material that failed to form
a satellite. In addition, many galaxies con-
tain ring-like structures. Consider a ho-
mogeneous thin ring of mass Mand
outer radius R(Fig. 13-52). (a) What
gravitational attraction does it exert on
a particle of mass mlocated on the
ring’s central axis a distance xfrom the
ring center? (b) Suppose the particle falls from rest as a result
of the attraction of the ring of matter. What is the speed with
which it passes through the center of the ring?
72 A typical neutron star may have a mass equal to that of the
Sun but a radius of only 10 km. (a) What is the gravitational accelera-
tion at the surface of such a star? (b) How fast would an object be
A
r
Earth
B
Figure 13-50
Problem 60.
R
r
M
P
Figure 13-51 Problem 68.
M
m
Rx
Figure 13-52
Problem 71.
384 CHAPTER 13 GRAVITATION
moving if it fell from rest through a distance of 1.0 m on such a star?
(Assume the star does not rotate.)
73 Figure 13-53 is a graph of the kinetic energy Kof an asteroid
versus its distance rfrom Earth’s center, as the asteroid falls di-
rectly in toward that center. (a) What is the (approximate) mass of
the asteroid? (b) What is its speed at r1.945 107m?
80 The fastest possible rate of rotation of a planet is that for which the
gravitational force on material at the equator just barely provides the
centripetal force needed for the rotation. (Why?) (a) Show that the
corresponding shortest period of rotation is
where ris the uniform density (mass per unit volume) of the
spherical planet. (b) Calculate the rotation period assuming a
density of 3.0 g/cm3, typical of many planets, satellites, and
asteroids. No astronomical object has ever been found to be
spinning with a period shorter than that determined by this
analysis.
81 In a double-star system, two stars of mass 3.0 1030 kg
SSM
TA3p
Gr,
K (10
9
J)
3
2
1
1.75 1.85
r (107 m)
1.95
Figure 13-53 Problem 73.
74 The mysterious visitor that appears in the enchanting
story The Little Prince was said to come from a planet that “was
scarcely any larger than a house!” Assume that the mass per unit
volume of the planet is about that of Earth and that the planet does
not appreciably spin. Approximate (a) the free-fall acceleration on
the planet’s surface and (b) the escape speed from the planet.
75 The masses and coordinates of three spheres are as
ILW
follows: 20 kg, x0.50 m, y1.0 m; 40 kg, x1.0 m, y1.0 m;
60 kg, x0m,y0.50 m.What is the magnitude of the gravita-
tional force on a 20 kg sphere located at the origin due to these
three spheres?
76 A very early, simple satellite consisted of an inflated
spherical aluminum balloon 30 m in diameter and of mass
20 kg. Suppose a meteor having a mass of 7.0 kg passes within 3.0
m of the surface of the satellite.What is the magnitude of the gravi-
tational force on the meteor from the satellite at the closest ap-
proach?
77 Four uniform spheres, with masses mA40 kg, mB35 kg,
SSM
mC200 kg, and mD50 kg, have (x, y) coordinates of (0, 50 cm),
(0, 0), (80 cm, 0), and (40 cm, 0), respectively. In unit-vector nota-
tion, what is the net gravitational force on sphere Bdue to the
other spheres?
78 (a) In Problem 77, remove sphere Aand calculate the gravi-
tational potential energy of the remaining three-particle system.
(b) If Ais then put back in place, is the potential energy of the
four-particle system more or less than that of the system in (a)?
(c) In (a), is the work done by you to remove Apositive or nega-
tive? (d) In (b), is the work done by
you to replace Apositive or negative?
79 A certain triple-star system
consists of two stars, each of mass m,re-
volving in the same circular orbit of ra-
dius raround a central star of mass M
(Fig. 13-54).The two orbiting stars are al-
ways at opposite ends of a diameter of
the orbit. Derive an expression for the
period of revolution of the stars.
SSM
M
m
m
r
Figure 13-54
Problem 79.
orbits a planet of mass M9.50 1025 kg, in a circular orbit of ra-
dius r4.20 107m. What are (a) the period of the orbit and (b)
the speed of the shuttle craft? Janeway briefly fires a forward-
pointing thruster, reducing her speed by 2.00%. Just then, what are
(c) the speed, (d) the kinetic energy, (e) the gravitational potential
energy, and (f) the mechanical energy of the shuttle craft? (g)
What is the semimajor axis of the elliptical orbit now taken by the
craft? (h) What is the difference between the period of the original
circular orbit and that of the new elliptical orbit? (i) Which orbit
has the smaller period?
84 Consider a pulsar, a collapsed star of extremely high density,
with a mass Mequal to that of the Sun (1.98 1030 kg), a radius R
of only 12 km, and a rotational period Tof 0.041 s. By what per-
centage does the free-fall acceleration gdiffer from the gravita-
tional acceleration agat the equator of this spherical star?
85 A projectile is fired vertically from Earth’s surface with an
initial speed of 10 km/s. Neglecting air drag, how far above the sur-
face of Earth will it go?
86 An object lying on Earth’s equator is accelerated (a) toward the
center of Earth because Earth rotates, (b) toward the Sun because
Earth revolves around the Sun in an almost circular orbit, and
(c) toward the center of our galaxy because the Sun moves around
the galactic center. For the latter, the period is 2.5 108y and the
radius is 2.2 1020 m. Calculate these three accelerations as multi-
ples of g9.8 m/s2.
87 (a) If the legendary apple of Newton could be released from
rest at a height of 2 m from the surface of a neutron star with a
mass 1.5 times that of our Sun and a radius of 20 km, what would be
the apple’s speed when it reached the surface of the star? (b) If the
apple could rest on the surface of the star,what would be the approxi-
mate difference between the gravitational acceleration at the top and
at the bottom of the apple? (Choose a reasonable size for an apple;
the answer indicates that an apple would never survive near a neu-
tron star.)
ILW
each rotate about the system’s center of mass at radius 1.0 1011 m.
(a) What is their common angular speed? (b) If a meteoroid
passes through the system’s center of mass perpendicular to
their orbital plane, what minimum speed must it have at the
center of mass if it is to escape to “infinity” from the two-star
system?
82 A satellite is in elliptical orbit with a period of 8.00 104s
about a planet of mass 7.00 1024 kg. At aphelion, at radius 4.5
107m, the satellite’s angular speed is 7.158 105rad/s. What is its
angular speed at perihelion?
83 In a shuttle craft of mass m3000 kg, Captain Janeway
SSM
385
PROBLEMS
nitrocellulose to a speed of 11 km/s along the gun barrel’s length of
220 m.(a) In gunits, what is the average acceleration of the capsule
and astronauts in the gun barrel? (b) Is that acceleration tolerable
or deadly to the astronauts?
A modern version of such gun-launched spacecraft (although
without passengers) has been proposed. In this modern version,
called the SHARP (Super High Altitude Research Project) gun,
ignition of methane and air shoves a piston along the gun’s tube,
compressing hydrogen gas that then launches a rocket. During this
launch, the rocket moves 3.5 km and reaches a speed of 7.0 km/s.
Once launched, the rocket can be fired to gain additional speed. (c)
In gunits, what would be the average acceleration of the rocket
within the launcher? (d) How much additional speed is needed
(via the rocket engine) if the rocket is to orbit Earth at an altitude
of 700 km?
97 An object of mass mis initially held in place at radial distance
r3REfrom the center of Earth, where REis the radius of Earth.
Let MEbe the mass of Earth. A force is applied to the object to
move it to a radial distance r4RE, where it again is held in place.
Calculate the work done by the applied force during the move by
integrating the force magnitude.
98 To alleviate the traffic congestion between two cities such as
Boston and Washington, D.C., engineers have proposed building a
rail tunnel along a chord line connecting the cities (Fig. 13-55). A
train, unpropelled by any engine and starting from rest, would fall
through the first half of the tunnel and then move up the second
half.Assuming Earth is a uniform sphere and ignoring air drag and
friction, find the city-to-city travel time.
closest and farthest distances are 1.47 108km and 1.52 108km
respectively. Determine the corresponding variations in (a) total
energy, (b) gravitational potential energy, (c) kinetic energy, and
(d) orbital speed. (Hint: Use conservation of energy and conserva-
tion of angular momentum.)
90 A 50 kg satellite circles planet Cruton every 6.0 h. The magni-
tude of the gravitational force exerted on the satellite by Cruton is
80 N. (a) What is the radius of the orbit? (b) What is the kinetic en-
ergy of the satellite? (c) What is the mass of planet Cruton?
91 We watch two identical astronomical bodies Aand B, each of
mass m, fall toward each other from rest because of the gravita-
tional force on each from the other. Their initial center-to-center
separation is Ri. Assume that we are in an inertial reference frame
that is stationary with respect to the center of mass of this two-
body system. Use the principle of conservation of mechanical
energy (KfUfKiUi) to find the following when the center-
to-center separation is 0.5Ri: (a) the total kinetic energy of the sys-
tem, (b) the kinetic energy of each body, (c) the speed of each body
relative to us, and (d) the speed of body Brelative to body A.
Next assume that we are in a reference frame attached to
body A(we ride on the body). Now we see body Bfall from rest to-
ward us. From this reference frame, again use KfUfKiUito
find the following when the center-to-center separation is 0.5Ri: (e)
the kinetic energy of body Band (f) the speed of body Brelative
to body A. (g) Why are the answers to (d) and (f) different? Which
answer is correct?
92 A 150.0 kg rocket moving radially outward from Earth has a
speed of 3.70 km/s when its engine shuts off 200 km above Earth’s
surface. (a) Assuming negligible air drag acts on the rocket, find
the rocket’s kinetic energy when the rocket is 1000 km above
Earth’s surface. (b) What maximum height above the surface is
reached by the rocket?
93 Planet Roton, with a mass of 7.0 1024 kg and a radius of 1600
km, gravitationally attracts a meteorite that is initially at rest relative
to the planet, at a distance great enough to take as infinite.The mete-
orite falls toward the planet. Assuming the planet is airless, find the
speed of the meteorite when it reaches the planet’s surface.
94 Two 20 kg spheres are fixed in place on a yaxis, one at
y0.40 m and the other at y0.40 m. A 10 kg ball is then
released from rest at a point on the xaxis that is at a great dis-
tance (effectively infinite) from the spheres. If the only forces
acting on the ball are the gravitational forces from the spheres,
then when the ball reaches the (x, y) point (0.30 m, 0), what are
(a) its kinetic energy and (b) the net force on it from the spheres,
in unit-vector notation?
95 Sphere Awith mass 80 kg is located at the origin of an xy coordi-
nate system; sphere Bwith mass 60 kg is located at coordinates
(0.25 m, 0); sphere Cwith mass 0.20 kg is located in the first quadrant
0.20 m from Aand 0.15 m from B. In unit-vector notation, what is the
gravitational force on Cdue to Aand B?
96 In his 1865 science fiction novel From the Earth to the
Moon, Jules Verne described how three astronauts are shot to the
Moon by means of a huge gun. According to Verne, the aluminum
capsule containing the astronauts is accelerated by ignition of
99 A thin rod with mass M5.00 kg is bent
in a semicircle of radius R0.650 m (Fig.13-56).
(a) What is its gravitational force (both magni-
tude and direction on a particle with mass
m3.0 103kg at P, the center of curva-
ture? (b) What would be the force on the parti-
cle if the rod were a complete circle?
100 In Fig. 13-57, identical blocks with identical
masses m2.00 kg hang from strings of different
lengths on a balance at Earth’s surface. The strings
have negligible mass and differ in length by h
5.00 cm. Assume Earth is spherical with a uniform
density r5.50 g/cm3.What is the difference in the
weight of the blocks due to one being closer to
Earth than the other?
101 A spaceship is on a straight-line path between Earth and the
Moon. At what distance from Earth is the net gravitational force
on the spaceship zero?
Train
Figure 13-55 Problem 98.
R
h
Figure 13-56
Problem 99.
Figure 13-57
Problem 100.
88 With what speed would mail pass through the center of Earth
if falling in a tunnel through the center?
89 The orbit of Earth around the Sun is almost circular: The
SSM
